An Object-Oriented preprocessor fit for C++

An Object-Oriented preprocessor fit for C++
An Object-Oriented preprocessor fit for C++

An Object-Oriented preprocessor fit for C++

E.D.Willink and V.B.Muchnick

Abstract: C++ retains the ANSI C preprocessor, although its limitations have been widely

recognised. We describe FOG, a meta-compiler for a super-set of C++, that provides

replacement preprocessing and introduces static meta-programming, while preserving the spirit

of C++. We show how implementation of preprocessor functionality in an Object-Oriented

style eliminates unnecessary replication from practical C++ programs, and supports recent

Object-Oriented Programming developments to a much greater extent than existing tools.

1 Introduction

Cpp, the C preprocessor, has always been an essential accompaniment for the C language. Apart from some minor rationalisation for the ANSI C standard [1], Cpp has survived unchanged as an important part of C++ [2]. Stroustrup, in [3], identifies elimination of the preprocessor as a major goal for C++, devoting the final chapter to a discussion of its weaknesses, and identifying some remedies that C++ provides. In the final paragraph, Stroustrup writes: “I’d like to see Cpp abolished. However the only realistic and responsible way of doing that is first to make it redundant, ...”. Alternative preprocessors such as m4 can be used for C++, but they also operate independently of the underlying language.

A preprocessor supports reconfiguration of source text at compile time using techniques such as file inclusion, conditional compilation and text replacement. Cheatham [4] identified three kinds of macro replacements:

?Text macros (text replaced by text - as exemplified by Cpp macros)

?Computational macros (text replaced by the result of a computation - as exemplified by inline functions and templates)

?Syntax macros (text replaced by the syntax tree representing a linguistically consistent construct)

The Flexible Object Generator (FOG) renders Cpp redundant, introducing syntax macros that integrate with C++ and solve the problems associated with text macros. Where Cpp performs lexical manipulation without regard to context, FOG provides extensive meta-level facilities and has a full understanding of C++ declarations.

Programmers exploit whatever tools are available to reuse ideas and avoid repetition: subroutines and classes can encapsulate some forms of reusable functionality; templates, particularly when used imaginatively, can provide reusable solutions to many more problems [5]. However when a problem of reuse is, or is perceived to be, insoluble within the programming language, programmers must resort to extra-lingual approaches. Lexical pasting with the preprocessor is inelegant and error prone, but to be preferred over abandoning reuse and replicating code.

Use of simple patterns [6] or idioms [7] leads to very predictable coding sequences. Many of these cannot be captured by a single C++ construct and so paradoxically C++ programs often use the preprocessor more than their C predecessors. Even with the assistance of the preprocessor, it is difficult to represent patterns that cut across multiple classes, and so the pattern used for design is lost from the implementation [8].

In this paper we show how a relatively simple subset of the FOG meta-level facilities may be used for practical applications without requiring an understanding of the underlying meta-concepts. Space does not permit more than a hint of how further use of the meta-level concepts can support weaving and Aspect-Oriented Programming

[9]. A description of the meta-concepts may be found in

[10], and an AOP example in [11].

We start by showing how the basic facilities of Cpp are replaced, using very simple examples, that are gradually reworked as more powerful facilities are described and exploited. We then examine a more typical example, before reviewing related work and relevance to other languages.

2 Traditional Preprocessing

2.1 Lexical substitution

Lexical substitution enables common definitions to be shared, given sensible names, and factored out if alternative definitions are needed in different contexts. When used responsibly, this leads to a considerable improvement in code quality, and is one of the main reasons for the widespread use of the preprocessor. However it is very easy for unfortunate substitutions to occur, and the presence of all names from all header files in a single name space is a source of many problems.

? IEE, 2000

IEE Proceedings online no. 20000573

DOI: 10.1049/ip-sen:20000573

Paper first received 17th September 1999 and in revised form 29th March 2000

E.D. Willink is with Racal Research Limited, Worton Drive, Reading RG2 0SB, UK

E-mail: ed.willink@https://www.360docs.net/doc/2610066210.html,

V.B. Muchnick is with the School of Electronic and Electrical Engineering, Information Technology and Mathematics, University of Surrey, Guildford GU2 5XH, UK

E-mail: v.muchnick@https://www.360docs.net/doc/2610066210.html,

C++ has removed the need for many substitutions by the

introduction of initialised const s and scoped enumerations. However, even where these are appropriate, the need for a non-integral type may defeat C++ enhancements.

Problems with Cpp substitution stem from the single namespace and from forceful substitution irrespective of context. Resolution of the namespace problem in FOG will be dealt with later. The problem of over-enthusiastic substitution is resolved by changing to a policy of substitution by invitation, rather than substitution by imposition. In FOG, instantiation of the definition of NAME is invited by $NAME, with the fallback of ${NAME} when subsequent characters could cause an unwanted meaning. The increased safety incurs the cost of the trigger characters to invite the substitution. These characters are not too out of place in a cryptic language such as C. The syntax should be familiar to Unix shell or make programmers.

2.2 Name concatenation

Name concatenation is useful for generating a new name derived from some stem. Thus an implementation of the NullObject pattern [12] may automatically define a Null class derived from its AbstractObject by suffixing Null to the class name of the AbstractObject.

class AbstractObject Null

: public AbstractObject

{ /*...*/ };

This can be realised directly in FOG, where unseparated identifiers and literals (numbers, strings and characters) are concatenated1.

class ${ABSTRACTOBJECT}Null

: public $ABSTRACTOBJECT

{ /*...*/ };

Cpp provides the ## concatenation operator that can only be used within a macro:

#define NULLOBJECT_INTERFACE(ABSTRACTOBJECT) \ class ABSTRACTOBJECT ## Null \

: public ABSTRACTOBJECT { /*...*/ };

2.3 String conversion

It is sometimes necessary, particularly for diagnostic purposes, to use a name as both an identifier and a string. const char *Class::class_name() const

{ return "Class"; }

This may be expressed directly in FOG, exploiting concatenation of an empty string to perform a lexical cast, since the result of a concatenation is of the same kind as the first contribution.

const char *${CLASS}::class_name() const

{ return ""$CLASS; }

Cpp provides the # operator for use within macros.

#define CLASS_NAME_IMPLEMENTATION(CLASS) \

const char *CLASS::class_name() const \ { return # CLASS; }

2.4 Text replacement

The preprocessor #define directive is used to define object-like macros

#define PI 3.14159

and function-like macros

#define max(a,b) ((a) > (b) ? (a) : (b)) supporting usage as

Obscure syntax incompatibilities requiring spaces in extern "C"and around and are described in [13] .The replacement text is an arbitrary sequence of preprocessor tokens that are substituted without regard to context. Errors, particularly in nested definitions, are difficult to diagnose, because substitution occurs before any language interpretation is applied; few compilers or debuggers support tracing back to the source once substitution has occurred. Long definitions require the use of backslashed continuation lines, which are inconvenient and unreliable to edit or read. Readability is further impaired by the need to use parentheses to guard against the possibility of accidental association problems.

FOG provides a meta-level where conventional run-time concepts can be used at (meta-)compile time. Meta-variables replace object-like macros and meta-functions replace function-like macros. Meta-variables and meta-functions are declared and typed in a very similar way to normal C++ variables and functions, save for the new use of the auto keyword and the introduction of meta-types: auto double PI = 3.14159; // Meta-variable

auto expression max(expression a, expression b) {// Meta-function $a > $b ? $a : $b;

}

for use as

a = $max(sin(2*$PI*f),0.5)

The auto keyword is almost totally obsolete in C++, where auto is only permitted within functions. auto is reused outside of functions in FOG to declare meta-functionality. Readers may choose to pronounce auto as meta, throughout this paper.

The meta-types correspond to the basic kinds of token (identifier, string and character), the numeric types (bool, double, int and etc.) and also to productions such as declaration and expression from the C++ grammar [2].

Use of meta-types enables the parser to ensure that arguments are passed and returned compatibly, and to diagnose errors more helpfully. When appropriate, conversions between the basic kinds are performed automatically.

Substitution within the meta-function replaces each invocation by its corresponding argument expression2.

The simple meta-function implementation of max solves the parenthesis problem, works for arbitrary types but remains prone to side effects. The invocation

$max(a++, b++)

will result in one argument receiving a double increment just as in Cpp.

2.5 Conditional compilation

Conditional compilation is essential to support a variety of configuration options, often to resolve distinctions between different operating systems. It may be appropriate to define static const char *temp_path = "/tmp/";

for use under Unix whereas NT might require

static const char *temp_path = "C:\\Temp\\"; FOG elevates C++ run-time statements such as if ... else ... for use at the meta-level, so that the selection may be made using an apparently conventional test:

auto if ($UNIX)

static const char *temp_path = "/tmp/";

else

static const char *temp_path = "C:\\Temp\\";

2Substitution is syntax-tree-based, rather than token-based as in the ANSI C preprocessor, or character-based as in the K&R preprocessor.

Cpp provides line-oriented conditional directives that mark-up rather than form part of the source text:

#if defined(UNIX)

static const char *temp_path = "/tmp/";

#else

static const char *temp_path = "C:\\Temp\\";

#endif

C++ statements occur only within functions. The use of the auto prefix is therefore redundant in the above example.

2.6 Other directives

A description of the replacements for other Cpp directives may be found in [13]. In summary, FOG provides a more integrated form of #error, a more disciplined context for #pragma and a form of #include that solves the problems of multiple inclusion. #line is not replaced.

3 Object-oriented preprocessing

The facilities described above provide consistent replacement for Cpp behaviour. Most of the extensions could be regarded as extensions to C rather than C++. Reviewing and generalising the facilities within the context of C++ leads to a much more powerful programming environment in which predictable program structures can be coded effectively.

3.1 Scopes

Meta-variables and meta-functions may be scoped and inherited, and meta-statements may occur within declaration scopes.

Revisiting the conditional compilation example of Section 2.5 from an Object-Oriented perspective, we find no need for conditional compilation. The characteristics of each configuration option may be packaged as meta-variables (and meta-functions) of a (meta-)class3.

class OsTraits_Abstract

{

auto bool NT = false;// default value

auto bool UNIX = false;

//...

};

class OsTraits_NT : public OsTraits_Abstract

{// derived class

auto bool NT = true;// overriding value

auto string temp_path = "C:\\Temp\\";

//...

};

class OsTraits_Unix : public OsTraits_Abstract

{

auto bool UNIX = true;

auto string temp_path = "/tmp/";

//...

};

OsTraits_NT may be configured as the implementation of OsTraits, by specifying the value of OS on the FOG command line

fog ... -D OS=NT ...

and using the built-in meta-function

auto string std::get_cpp(string macroName)

to access it from

class OsTraits

: public OsTraits_$std::get_cpp("OS") {}; thereby creating the equivalent declaration

class OsTraits : public OsTraits_NT {};

This maps the required configuration to OsTraits, so

In FOG, every class and built-in type has an identically named meta-class, so for the purposes of this paper classes and meta-classes may be considered equivalent.that an operating system specific file may be defined using the temporary path by

const char *fileName =

$OsTraits::temp_path "results.dat";

This is then resolved at compile-time to

const char *fileName = "C:\\Temp\\results.dat"; Having isolated the configuration in separate classes and an associated header file, a new operating system can be supported by providing a prefix file characterising the new system and invoking it with an appropriate command line. Existing source files need no change. This could be achieved directly using multiple layers of name substitutions with C preprocessor, but it never is. Modularisation is much easier when supported by the programming environment. This cannot be achieved using C++ templates, which lack the ability to perform string manipulations.

3.2 Compilation model

C++ supports a two stage translation process involving multiple independent compilations followed by a link editing stage to produce a final executable. The independent compilations are consistent provided the One Definition Rule [2 (§3.2)] is observed. Simply stated, this rule requires that a declaration in one compilation must not have a different meaning in any other. In practice, placing declarations in header (interface) files, which are included by each compilation session that requires them, usually satisfies the One Definition Rule.

From the perspective of a compiler writer, the One Definition Rule is very useful, if not essential. From the perspective of the programmer, the One Definition Rule is very inconvenient. Declarations must be provided twice, once in the interface file and again in the implementation file. Declarations cannot be freely interleaved. In more serious applications, a conflict arises between language constraints and the programmer’s need to organise code to suit algorithmic or functional perspectives. Code has to be organised to suit the compiler. Patterns cannot be preserved in the code [8] and Aspect-Oriented Programming [9] is not readily supported.

A preprocessor for C++, that performs its processing prior to compilation, can bridge the gap between the organisational requirements of the programmer and the integrity requirements of the compiler. FOG operates in this way using an augmented compilation model as shown in Fig. 1.

The centre and right hand sides show the conventional C++ compilation model. Interface files provide the declarations to be shared by independent compilations, which produce object files to be linked together with libraries to produce an executable. (The complexities of static construction and template instantiation are conveniently hidden by the ‘Linker2’ activity.) Meta-compilation adds the extra stages on the left hand side. The conventional C++ interface and implementation files are generated by one or more meta-compilations from source files (the forward arrows) and from frozen interfaces (the reverse arrows). Sources may be shared between meta-compilations, and a single meta-compilation may generate any number of interfaces and/or implementations.

Clearly the One Definition Rule must still be respected by the interface and implementation files fed to the compiler. However a more relaxed Composite Definition Rule can now be imposed on the source files. Simply stated, the composite meaning of all like declarations must

meaning is explained in Section 3.4.

3.3 Joint interface and implementation Introduction of a meta-compiler that synthesises interface

and implementation files eliminates the need for independent interface and implementation declarations. It is appropriate to generalise C++ declarations to remove the distinction between interface-specific and implementation-specific declarations. This generalisation turns out to be almost entirely semantic, since the C++ grammar already permits an interface-specific keyword such as virtual to accompany a function-definition4. It is only necessary to allow an access-specifier(e.g. protected) as part of a decl-specifier (the type part of a declaration), and to permit a full id-expression(e.g. Scope::name) where previously only an identifier was allowed.

Programmers may then use an implementation style of declaration for parts of interfaces

public typedef size_t Class::SizeType;

or provide complete implementations in interfaces:

class Class

{

protected virtual void f(int x = 0) = 0

{ std::cout << x; }

public:

static double y = 0;

};

A complete solution to the class_name()example from Section 2.3 may now be captured by the single meta-function

auto declaration ClassName()

{

public virtual !inline

const char *class_name() const

{ return ""$Scope; }

};

which can be invoked as

class NamedClass

{

$ClassName();

};

The reserved meta-variable Scope refers to the prevailing scope, avoiding the need to pass it as a parameter.

The italicised terms correspond to productions in the C++ grammar [2].function body is not inlined. Similarly !static would provide for explicit rather than default programming intent.

The single meta-function invocation generates the equivalent C++ interface

class NamedClass

{

public:

virtual const char *class_name() const;

};

and implementation

const char *NamedClass::class_name() const { return "NamedClass"; }

This requires a pair of macros when implemented using Cpp.

#define CLASS_NAME_INTERFACE() \

virtual const char *class_name() const;

#define CLASS_NAME_IMPLEMENTATION(CLASS) \

const char *CLASS::class_name() const \

{ return # CLASS; }

and a corresponding pair of invocations one from the interface

CLASS_NAME_INTERFACE()

and one from the implementation

CLASS_NAME_IMPLEMENTATION(NamedClass)

3.4 Composition

In C++, multiple declarations are an error. In FOG, multiple compatible declarations are composed; only incompatible declarations are an error. Space does not permit more than a superficial exposition of the composition rules.

Composed declarations merge their components, and so a variable qualified with static carries the static with it when merged with another variable that has no static specification, but provokes an error message if merged with a !static.

Overloaded function declarations compose independently. Default values may be repeated but may not conflict.

Arrays and enumerations extend to accommodate all contributions. Duplicate initialisations must match. Holes

in arrays are zero filled. The GNU C [14] extension is supported so that sparse arrays can be defined and composed.

bool is_prime[] = { [2] true, true, [5] true,

[7] true, [11] true };

The constructor initialisers for a particular constructor are composed and must not conflict. Unspecified initialisers for non-copy constructors are obtained from member variable initialisers. For example, code to support an error handling aspect may add a member variable with a default initialiser:

public bool Class::_error_generated = false;

A constructor independently added in support of some other aspect

Class::Class(PersistenceManager&) /*...*/; provides the requisite initialisation.

Classes expand to encompass all distinct member declarations, with repeated declarations composed recursively.

Function (and constructor) bodies are composed by concatenating code contributions within named regions, which are in turn concatenated to form the overall function body. The regions named entry and exit typically provide for variable declaration and initialisation and a return statement, ensuring a predictable structure. Regions named pre and post provide code to operate before or after the default body region of the function. Function definitions are extended to support a declarative scope within which regions are prefixed by their name.

public bool Manager::do_it()

:{// Start of declarative scope

entry { bool exitStatus = true; };

exit { return exitStatus; };

};

defines a framework for a composed function. A return variable is initialised in the entry region, and returned by the exit region. With the framework in place, code concerned with a particular aspect may contribute code to the function:

private Aspect Manager::_aspect;

public bool Manager::do_it()

{

if (!_aspect.do_it())

exitStatus = false;

}

FOG weaves the contributions together to produce the equivalent C++ declarations:

class Manager

{

private:

Aspect _aspect;

public:

bool do_it();

};

bool Manager::do_it()

{

bool exitStatus = true;

if (!_aspect.do_it())

exitStatus = false;

return exitStatus;

}

Readers who have programmed extensively with a macro assembler may recognise that the ability to extend classes, function code regions, enumerations and arrays at will gives each declaration space the attributes of a program section.

It is possible to define a meta-function that performs extension of an enumeration and a text array so that

5 A [constant-expression] preceding an array initializer specifies the array index to be initialized.numeric and text declarations are automatically synchronised.

auto declaration NamedEnum(identifier aName)

{

public enum Enum { $aName };

public static const char *names[] =

{ ""$aName };

}

Invocation as

class Colours

{

$NamedEnum(RED);

$NamedEnum(GREEN);

$NamedEnum(BLUE);

};

provides successive entries for Colours::Enum and corresponding entries for Colours::names[], as if the user had typed:

class Colours

{

public:

enum Enum { RED, GREEN, BLUE };

static const char *names[];

};

and

const char *Colours::names[] =

{ "RED", "GREEN", "BLUE" };

The conversion of a single name such as RED into multiple interleaved declarations cannot generally be achieved using the preprocessor or C++ templates.

3.5 Derivation rules

There are many idioms that require entirely predictable code to be provided by derived classes in order to comply with a protocol defined by a base class. The class_name()method of Section 3.3 provides one example. In C++, a declaration applies to the scope for which it is specified. In FOG, this scope is referred to as the root scope for that declaration. An optional derivation rule specifies how that declaration may be automatically redefined in the inheritance tree of scopes that derive from the root scope. Refining the example from Section 3.3

auto declaration ClassName()

{

public virtual !inline

const char *class_name() const

:{

derived(true) { return ""@Scope; };

};

};

A declarative scope has been introduced to prefix a derivation rule to the function body. The predicate of derived(true) is always true and so the declaration is applied throughout the entire inheritance tree, that is at the root scope and all derived scopes.

The change of substitution operator from $to @, changes the evaluation time. $is an early substitution operator, evaluated when source tokens are first parsed to create a potential declaration in its associated root scope, at which point Scope resolves to the root scope. @ is a late substitution operator, evaluated when a potential declaration becomes an actual declaration in its eventual scope, at which point Scope resolves to the actual scope. (If the $operator were used in the example, all derived scopes would return the name of the root scope.) Derivation rules can apply to the declaration of any entity. Michael Tiemann provided a solution [3] to the problem of providing a mnemonic name for the primary base class

typedef employee inherited;

//...

void print();

};

class manager : public foreman {

typedef foreman inherited;

//...

void print();

};

enabling a derived class to refer to its base class mnemonically as inherited rather than explicitly.

void manager::print()

{

inherited::print();

//...

}

In FOG, the entire hierarchy of typedef s can be expressed by a single declaration.

private typedef @Super employee::inherited

:{ derived(!is_root()); };

This provides a typedef for all derived classes. The derivation predicate inhibits the declaration at the root, where the built-in meta-variable Super may have no valid resolution for the primary base class.

The Prototype pattern [6], virtual constructor, or cloning idiom [15] is also provided very easily using a derivation rule. The conventional approach requires that a clone method be defined for every non-abstract class in an inheritance hierarchy

class ConcreteClass /* ... */

{

/* ... */

virtual RootClass *clone() const;

};

RootClass *ConcreteClass::clone() const

{ return new ConcreteClass(*this); }

This requires the programmer to manually weave the code in to every class. This is potentially error prone and costs at least one line per interface and one line per implementation file of every class. Using FOG, the idiom can be fully defined by a meta-function:

auto declaration Prototype()

{

public virtual $Scope *clone() const = 0

:{

derived(!is_pure())

{ return new @{Scope}(*this); };

};

}

The !is_pure()derivation predicate specifies that the declaration contributes code to all derived classes that have no pure virtual functions.

Instantiation requires a single line in the base class that defines the protocol. No code is required in derived classes.

class Base

{

$Prototype();

};

($Scope may be changed to @Scope to use the derived type as the return type.)

These two examples demonstrate FOG at its most advantageous: one line in the base class guarantees protocol observance and replaces one or more lines in each derived class. A more realistic example will now be examined.

4 A Real Example

One of the activities of a compiler involves selection of appropriate machine instructions (such as ADD or MOVE)to implement the program, usually represented by a tree of Abstract Syntax Tree nodes [16]. An effective approach to solving this problem involves a Bottom-Up Rewrite System [17], which searches the tree from the leaves upwards identifying the lowest cost solution that has each node covered exactly once by a machine instruction. The tree may then be rewritten in terms of the selected machine instructions. In order to support multiple target architectures, alternative instruction sets must be supported. Implementation of this diversity is assisted by the use of a Bottom-Up Rewrite Generator to transform a description of each machine instruction into the form needed for an efficient tree search. An example of this form of generator is lburg that forms part of the lcc C compiler [18].

lburg is a compact C program comprising just three files. lburg.c has 690 lines and 4652 (non-comment, non-whitespace preprocessor) tokens. lburg.h has 66 lines and 259 tokens. Additionally gram.y is a 19 rule, 37 state yacc parser grammar.

lburg supports single dispatch architectures (such as SPARC). An enhanced version was required in order to support less conventional processor architectures, and so a highly Object Oriented C++ rewrite was undertaken using reference counting and smart pointers to share common partial instructions. The resulting program was substantially larger, due to the extra declarations for encapsulated C++ classes, rather than the original free access to structure elements, and due to the added functionality. Preprocessor macros were used extensively to factor out common declarations.

A further revision to exploit FOG without any other change of functionality forms the basis of the following comparison. An implementation based on the use of preprocessor macros is compared with an implementation using meta-functions, meta-variables and derivation rules.

Use of FOG reduced the token count by 14%, from 20250 to 17500. The per-class reduction varied between 9 and 48%. The larger reductions occurred in small classes, where the benefits of derivation rules and simplification of interface and implementation declarations were most apparent.

A reduction in token count is an easily measured reduction in programming effort. Less easily measured are the more aesthetic improvements of better modularity, improved expression of programming intent, and automatic compliance with programming protocols. A pair of short before/after extracts are provided in the Appendix for readers to make their own judgements. The code is complete save for the removal of 4 functions whose lexical structure exactly duplicates functions that remain. Code for this example is chosen because it is shortest, and so demonstrates the changes more clearly. Providing the large number of unaffected function body lines from a more typical module would not provide extra insight. Space does not permit the definitions of the preprocessor macros or meta-functions to be shown. The two are of comparable lexical size, the meta-function has a higher token count through the use of $ operators and meta-type names, but a lower token count through the use of more appropriate facilities. The meta-functions are modular, having fewer interdependencies than the preprocessor macros, and more readable through the use of more conventional structuring and the elimination of back-slash continuation lines.

The original preprocessor macros almost vanish completely. The CUSTOM_RTTI support is provided automatically by derivation. The remaining six macros supporting smart pointers are all subsumed by

MapOfSmartPointerSpecialisations. Other meta-functions such as Mutate just implement simple idioms.

5 Related Work

Although extensively criticised [3], relatively little work has been published on alternatives to the C preprocessor [1]. Weise [19] provides a review of earlier work, and describes the use of a syntactic macro, which has fully-typed argument and return values. Invocation of the macro occurs within the context of a parse tree, so there is no opportunity for the unpleasant side effects that occur with conventional macros that lack sufficient parentheses. The usage of fully meta-typed arguments for meta-functions in FOG is influenced by Weise, however the FOG notation is more compact and supports both character- and syntax-based replacement. Weise introduces nine extra lexical operators, requires an explicit return, and produces examples that are unpleasant and sometimes difficult to read. FOG introduces only two extra lexical operators ($ and @) and by treating the entire meta-function body as the return achieves a simpler substitution model. Weise considers only ANSI C, whereas FOG revisits the concepts with an Object-Oriented and meta-level perspective.

Researchers in many fields have chosen to use C++, but found it inadequate for their purposes. There are therefore many domain specific extensions to C++, just some of which are discussed below.

A low level understanding of object layout is necessary for persistent storage of objects in databases or for marshalling objects, whether for signalling between nodes in a communication network, or distribution across nodes in a parallel processor. Wilson and Lu [20] provides extended articles by 16 of the leading research teams using C++ for parallel processing. Some researchers used only library classes and run-time support code, and so remain entirely within the normal confines of the C++ language. Others introduce language extensions, which are variously implemented as translators to C++, or modified C++ compilers. MPC++ [21] exploits meta-level facilities to support an extended syntax within a ‘standard’ C++ compiler. Many of the C++ extensions appear unnecessary and some authors recognise that more imaginative use of C++ facilities, particularly those not readily available at the start of their research could have reduced the need for divergence.

Domain specific extensions, when fully integrated with C++, can provide a clean solution to the domain problem. Many extensions are poorly integrated because of the size and complexity of C++ and so provide little more than a research tool. Many of the problems dealt with in a domain specific fashion can be resolved in a domain independent fashion by using the meta-level programming facilities of FOG. However FOG meta-programming is restricted to declarations and so the more radical changes of C** [22] in which data parallel semantics are introduced to expressions could certainly not be addressed.

The concepts of meta-classes were first defined for Smalltalk. Languages such as CLOS have been extended with a Meta-Object Protocol (MOP) [23]. Even Java has a meta-class object for every class. C++ has rather lagged behind, perhaps through a mismatch of the run-time characteristics of traditional MOPs and the statically compiled philosophy of C++, perhaps through the compiler writer’s desire to prevent further explosion of language complexity. FOG provides statically compiled meta-functionality, which can be used to define customised run-time meta-functionality. It is difficult to answer the critique that C++ is too large, and that adding meta-functionality is an enhancement too far. However it is also difficult to avoid recognising that the absence of meta-functionality is restrictive for some domains and an inhibition to reuse for all.

C++ has no compile-time meta-level capabilities. Chiba [24] describes an extended form of C++ called Open C++, that allows library developers to write code to analyse class layout and so ensure that persistence can be resolved transparently. Meta-classes are used as adjuncts of the normal C++ class structures, with a wide variety of reserved meta-functions available for re-implementation to enable the parse tree to be adjusted during the compilation process. This provides considerable power, but because Open C++ programming is very closely related to the compiler internals, this approach is not suitable for normal programming. FOG also uses meta-classes, but every C++ class or built-in type is a FOG meta-class, and the meta-code creates declarations by using declaration statements directly, with the result that the facilities available at the meta-level in FOG are very similar in syntax and behaviour to those already available in C++. FOG is not able to rewrite expressions in the same way that Open C++ does, but there may be no need to. Relevant expression terms can be encapsulated within inline functions and templates, which FOG provides the ability to manipulate at compile-time.

The use of patterns has provoked considerable interest since the original patterns book [6], although most attempts to represent patterns in code are informal. Soukup addresses the problems of implementing patterns in [8].

Aspect-Oriented Programming [9] has recognised that an aspect may cut across many objects, requiring the contributions to each class from each aspect to be woven together.

Generative Programming [25] seeks to configure re-usable components, making extensive use of C++ templates to perform the compile-time configuration. This approach works well for components that can be fully defined by a single type or function. FOG complements the template approach by providing a compile-time flexibility to configure multiple declarations as well as single types or functions.

6 Other languages

This paper has concentrated on C++, where meta-level facilities are very limited and preprocessing is more extensively used than in other languages. In C++, an efficient program is produced at compile-time avoiding the need for run-time activities. Preprocessing contributes to resolving problems at compile-time. The compile-time processing available from static meta-programming is compatible with the C++ philosophy. It is less relevant and probably unwelcome for languages such as Smalltalk or CLOS that have a tradition of run-time flexibility.

Java has its origins in C++, and also seeks to resolve problems at compile-time although run-time performance is less critical. The absence of a preprocessor in Java eliminates one of the programmer’s options. The lexical and meta-level concepts in this paper could certainly be applied to Java, where the cleaner syntax and existing meta-classes could be exploited.

Significant revision of the syntax is necessary to make the concepts compatible with languages such as Ada 95 or Eiffel, for which some form of meta-level programming

would be beneficial, although a cryptic C/C++ lexical style would be out of place.

7 Performance

FOG is available as source, NT and Solaris binaries from [26]

The current implementation of FOG is at the level of a research tool and so the efficiency falls well below that of a production compiler.

The potential efficiency depends on the way in which the meta-compiler is used. If used to meta-compile on a class by class basis, the need to understand the context of declarations may incur a significant penalty, similar to the costs of instantiating templates one at a time without precompiled headers. If used on a subsystem by subsystem basis, the costs are amortised and can be similar to those of a conventional compiler.

A meta-compiler operating as an independent preprocessor duplicates much of the parsing and semantic analysis of the subsequent compiler. Using an appropriate intermediate representation to communicate between meta-compilation and compilation can reduce the meta-compilation overhead.

8 Conclusion

A meta-compiler has been described that makes the C preprocessor redundant

?concatenation and substitution replace # and ##?meta-variables and meta-functions replace object-like and function-like macros

?meta-statements replace #if

The meta-compiler extends C++ but preserves its essential characteristics

?costs are incurred only at compile-time

?syntax-based substitution respects program structure ?meta-types define meaning and support error diagnosis ?meta-classes define the scope of meta-variables and meta-functions

?meta-statements support meta-programming at compile-time

?derivation rules support and enforce inheritance hierarchy protocols

Although each extension is not new to computer science, it is their combination that provides the unique ability to avoid repetition in practical examples such as cloning.

Finally the composition of declarations frees the programmer from the constraints imposed by the One Definition Rule, providing the flexibility to encapsulate solutions to patterns as meta-functions and to weave together the declarations from each separate concern of an Aspect Oriented Program.

9 Acknowledgements

The authors thank an anonymous referee for suggesting a clearer exposition and Matthew Ellis, Brian Hillam and Mike Quinlan for reading drafts diligently. The first author is very grateful to Racal Research Ltd. for providing the time and resources to pursue this research.

10 References

1ANSI X3.159-1989.: ‘American National Standard for Information Systems -Programming Language - C’. (American National Standards Institute, 1990) 2ISO/IEC 14882:1998(E).: ‘International Standard - Programming Languages -C++’. (American National Standards Institute, 1998)

3STROUSTRUP, B.: ‘The design and evolution of C++’. (Addison-Wesley, 1994)

4CHEATHAM, T.E.: ‘The introduction of definitional facilities into higher level programming languages’. Proceedings of the 1966 Fall Joint Computer Conference, AFIPS, 1966, 29. pp. 623-637

5VELDHUIZEN, T.: ‘Using C++ template metaprograms’. C++ Report, 1995, 7, (4), pp. 36-43. (https://www.360docs.net/doc/2610066210.html,/~tveldhui/papers/meta-art.ps)

6GAMMA, E., HELM, R., JOHNSON, R., and VLISSIDES, J.: ‘Design patterns, elements of reusable object-oriented software’. (Addison-Wesley, 1995)

7COPLIEN, J.O.: ‘Advanced C++ programming styles and idioms’. (Addison-Wesley, 1992)

8SOUKUP, J.: ‘Taming C++: Pattern classes and persistence for large projects’. (Addison-Wesley, 1994)

9KICZALES, G., LAMPING, J., MENDHEKAR, A., MAEDA, C., LOPES,

C., LOINGTIER, J-M., and IRWIN, J.: ‘Aspect-Oriented Programming’.

Proceedings of the 11th European Conference on Object-Oriented Programming, ECOOP’97, June 1997, (Springer-Verlag, LNCS 1241)

(https://www.360docs.net/doc/2610066210.html,/spl/groups/eca/pubs/papers/Kiczales-

ECOOP97/for-web.pdf)

10WILLINK, E.D., and MUCHNICK, V.B.: ‘Preprocessing C++: Meta-class aspects’. Proceedings of the Eastern European Conference on Technology of Object Oriented Languages and Systems, June 1999, Blagoevgrad, Bulgaria, pp. 136-149, (ISBN 954-90484-1-1).

https://www.360docs.net/doc/2610066210.html,/Research/CSRG/fog/FogToolsEE2.pdf

11WILLINK, E.D., and MUCHNICK, V.B.: ‘Weaving a way past the C++ one definition rule’. Position paper for the Aspect-Oriented Programming Workshop at ECOOP’99, June 1999.

(https://www.360docs.net/doc/2610066210.html,/Research/CSRG/fog/AopEcoop99.pdf)

12MARTIN, R.C., RIEHLE, D., and BUSCHMANN, F.: ‘Pattern languages of program design 3’. (Addison-Wesley, 1997)

13WILLINK, E.D.: ‘Meta-compilation for C++’. PhD Thesis, Computer Science Research Group, University of Surrey, January 2000

(https://www.360docs.net/doc/2610066210.html,/Research/CSRG/fog/FogThesis.pdf )

14STALLMAN, R.M.: ‘Using and porting GNU C’. (Free Software Foundation, Inc., Cambridge MA, 1998. Available as part of the gcc-2.8.0 distribution.) 15STROUSTRUP, B.: ‘The C++ programming language’. (Addison-Wesley, 1997) 3rd edn.

16AHO, A.V., SETHI, R., and ULLMAN, J.D.: ‘Compilers: Principles, techniques and tools’. (Addison-Wesley, 1986)

17PROEBSTING, T.A.: ‘BURS Automata generation’. ACM Transactions on Programming Languages and Systems, 1995, 17, (3), pp. 461-486

18FRASER, C.W., and HANSON, D.R.: ‘A retargetable C compiler: Design and implementation’. (Benjamin/Cummings 1995)

19WEISE, D., and CREW, R.: ‘Programmable syntax macros’. Proceedings of the ACM SIGPLAN’93 Conference on Programming Language Design and Implementation, ACM SIGPLAN Notices, 1993, 28, (6), pp. 156-165

20WILSON, G.V., and LU, P.: ‘Parallel Programming using C++’. (MIT Press, 1996)

21ISHIKAWA, Y., HORI, A., TEZUKA, H., MATSUDA, M., KONAKA, H., MAEDA, M., TOMOKIYO, T., NOLTE, J., and SATO, M.: ‘MPC++’, in WILSON, G.V., and LU, P.: ‘Parallel Programming using C++’. (MIT Press, 1996), pp. 429-464

22LARUS, J.R., RICHARDS, B., and VISWANATHAN, G.: ‘C**’, in WILSON, G.V., and LU, P.: ‘Parallel Programming using C++’. (MIT Press, 1996), pp. 297-342

23KICZALES, G., DES RIVIéRES, J., and BOBROW, D.G.: ‘The art of the metaobject protocol’. (MIT Press, 1991)

24CHIBA, S.: ‘A metaobject protocol for C++’. Proceedings of the 1995 ACM SIGPLAN Conference on Object-Oriented Programming Systems, Languages and Applications, OOPSLA’95, ACM SIGPLAN Notices, 1995, 30, (10), pp.

285-299

25CZARNECKI, K., and EISENECKER, U.W.: ‘Synthesizing objects’.

Proceedings of the 13th European Conference on Object-Oriented Programming ECOOP’99, June 1999, Lisbon, Portugal, pp. 18-42, (Springer, LNCS 1628)

26https://www.360docs.net/doc/2610066210.html,/Research/CSRG/fog

11 Appendix

11.1 Original interface file

#ifndef ENTRY_HXX

#define ENTRY_HXX

#include

#include // A smart string class

#include

#include

#include

class Entry : public Object

{

CUSTOM_RTTI_DECLARATION(Entry, Object)

REFERENCE_COUNT_DECLARATION(Entry)

NULL_OBJECT_DECLARATION(Entry)

private:

const Burg& _burg;

const IdHandle _id;// Handle for a smart string

private:

Entry(const Entry&);// No copy

Entry& operator=(const Entry&); // No assign

protected:

Entry();

Entry(Burg& aBurg, const Id& anId);

public:

const Burg& burg() const { return _burg; }

const Id& id() const { return *_id; }

virtual Term *is_term();

const Term *is_term() const { return ((Entry *)this)->is_term(); }

virtual void mark_reachable();

virtual ostream& print_this(ostream& s) const;

};

#endif

11.2 Original implementation file

#include

#include

#include

CUSTOM_RTTI_IMPLEMENTATION(Entry, Object)

REFERENCE_COUNT_IMPLEMENTATION(Entry)

NULL_OBJECT_IMPLEMENTATION(Entry)

SMART_POINTER_IMPLEMENTATION(Entry)

MAP_OF_SMART_POINTER_IMPLEMENTATION(Entry)

Entry::Entry() : _burg(Burg::null_object()) {}

Entry::Entry(Burg& aBurg, const Id& anId) : _burg(aBurg), _id(anId) { aBurg.add_entry(*this); } Term *Entry::is_term() { return 0; }

void Entry::mark_reachable() {}

ostream& Entry::print_this(ostream& s) const { return s << _id; }

11.3 Revised FOG code, with use of FOG extensions italicised

using "Burg.fog";// Improved form of #include.

class Entry : public Object

{

using/interface "Burg.h"; // Need a #include

$NoCopy();// Entry(const Entry&);

$NoAssign();// Entry& operator= (const Entry&)

$Mutate();// Entry& mutate() const { return *(Entry *)this; } private:

const Burg& _burg = Burg::null_object();

const IdHandle _id;

protected:

!inline Entry() {}// Uses default initialiser value

public:

const Burg& burg() const { return _burg; }

const Id& id() const { return *_id; }

virtual Term *is_term() { return 0; }

const Term *is_term() const { return mutate().is_term(); }

virtual void mark_reachable() {}

virtual ostream& print_this(ostream& s) const { return s << _id; }

};

$MapOfSmartPointerSpecialisations(Entry);

protected Entry::Entry(Burg& aBurg, const Id& anId) : _burg(aBurg), _id(anId) { aBurg.add_entry(*this); }

最新人教版小学三年级(下)数学解决问题200道

最新人教版小学三年级(下)数学解决问题200道 16、我校三年级⑴班有4个小组,每个小组有9人,他们在植树节共植树180棵。平均每人植树多少棵 17、将牛奶倒入同一个空瓶里。如果倒入4杯牛奶,则连瓶重375克,如果倒入6杯牛奶,则连瓶重535克,算一算这个空瓶重多少克 18、一本软面抄元,一支钢笔元。买一本软面抄和一支钢笔一共要付多少元 19、63名少先队员去历史博物馆参观,租3辆客车,平均每辆坐多少人 20、一块菜地的长是20米,宽是15米,这块菜地的面积是多少平方米 21、小飞爸爸的体重是84千克,爸爸的体重是妈妈的2倍,是小飞的4倍。小飞妈妈的体重是多少千克小飞的体重是多少千克 22、林叔叔一天卖完了8箱冰棍,每箱冰棍有50根,每根2元。林叔叔一共卖了多少钱 , 23、电视台打算在暑假为小朋友们播出动画片《百变马丁》。共54集。 ⑴每天播出一集,至少要几个星期才能播完 ⑵如果只在每星期的周一至周六播出一集,需要几星期才能报完 24、有750粒棒棒糖,分给5组,每组有5人。 ⑴平均每个组分到几朵花 |

⑵平均每人分到几朵花 25、小富买8本一样的笔记本,要给老板80元,一个笔记本多少元小富觉得太贵,想换种便宜点的,老板介绍了一种7本77元的,老板说的话可信吗为什么 26、三年级⑴班有学生48人,每两人用一张课桌,一共要用多少张课桌把这些课桌平均摆成3列,每列摆多少张 ~ 27、为了吸引更多的游客,对某市世界之窗的门票,各个旅行社都亮出了自己的底价。小荣和小华选择了78元2人(只能玩其中几个项目)的套餐价,如果要玩遍所有项目,每人需要另外各交42元,那么小荣玩遍所有项目需要多少钱 28、一支铅笔元,比一支水笔便宜元,买一支铅笔和一支水笔一共要付多少元 29、青龙小学的学生到电影院看电影,电影院共有18排座位,每排24个。现在有学生418个,够坐吗如果够,多多少个座位如果不够,少多少个座位 30、光华小学准备为腰鼓队购买服装,下面是队员们的身高记录。(单位:厘米) 女生:126 134 124 132 144 130 127 127 128 130 - 男生:137 128 134 141 130 141 137 126 132 126 请你将队员们的身高统计在一张复式统计表里。 31、一艘货船7小时航行210千米,照这样的速度,这艘货船一天可航行多少千米 32、一场篮球赛从上午9时30分开始,进行了120分,比赛结束时的时间用24时计时法表示是多少

小学数学竞赛试题

小学数学创新能力竞赛(预赛)试题 一、填空题(每空3分,共60分) 1.20500321000≈()亿37094000=()万 2.甲比乙多20%,乙比甲少()。 3.用2、0、0、6可以组成()个不同的四位数。 4.能被2、3、5、7整除的三位数中,最大的是()。 5.用2、6、8和4个零组成的7位数中,只读出一个零的最大的数是()。 6.同学们排队从学校出发去看电影,队伍全长200米,从排头出校门到排尾进入电影院共用35分钟,如果步行的平均速度是每分钟50米,学校到电影院共()米。 7.找规律填得数:2.5 1.250.625()0.15625。 8.2006年世界杯足球赛分为8个小组,每组4支球队,每组进行循环赛(即:每支球队都与其它球队进行一场比赛),循环赛后每组选2支球队进行淘汰赛(即:每支球队进行一场比赛,赢的进入下一轮,输的淘汰),最后决出冠军。这次世界杯一共举行()场足球赛。 9.在1~2006这2006个自然数中,不能同时被7和13整除的数共 有()个。 10.如图1,大圆内画一个最大的正方形,正方形内画一个最大的 圆……,如此画下去,共画了4个圆,那么最大的圆的面积是最 小圆的()倍。 11.学校门口到公路边有一条100米的路,如果在这条路的两边栽树,离校门口10米处栽一棵,然后每隔10米栽一棵,一共需要栽()棵。 12.在方框里填上适当的数:50.15×[72.05- -17.95)]=2006 13.用88个小正方体表面积之和的比是()。 14.请你用1~9这九个数字,写出五个平方数(某个数的平方),每个数字最多用一次,这五个平方数分别是()。 15.2005年12月8日是星期四,推算一下,2006年5月1日是星期()。16.一个池塘里的睡莲,每天增长一倍,到第5天已长满了整个池塘,第二天长到这个池塘的()。 17.五年级参加植树活动,人数在30与50人之间,如果分3人一组,4人一组,6人一组或8人一组,都恰好分完。五年级参加植树的学生有()人。 图1

小学高段趣味数学100题附答案

小学高年级数学趣味题 1.8个数字“8”,如何使它等于1000 答案:8+8+8+88+888 2.小强数学只差6分就及格,小明数学也只差6分就及格了,但小明和小强的分数不一样, 为什么 答案:一个是54分,一个是0分 3.一口井7米深,有只蜗牛从井底往上爬,白天爬3米,晚上往下坠2米。问蜗牛几天能 从井里爬出来 答案:5天 4.某人花19快钱买了个玩具,20快钱卖出去。他觉得不划算,又花21快钱买进,22快 钱卖出去。请问它赚了多少钱 答案:2元 5.100个包子,100个人吃,1个大人吃3个,3个小孩吃1个,多少个大人和多少小孩刚 好能吃完 答案:25个大人,75个小孩 6.小王去网吧开会员卡,开卡要20元,小王没找到零钱,就给了网管一张50的,网管找 回30元给小王后,小王找到20元零的,给网管20元后,网管把先前的50元还给了他, 请问谁亏了 答案:网管亏了30元 7.每隔1分钟放1炮,10分钟共放多少炮 答案:11炮 8.一个数去掉首位是13,去掉末位是40.请问这个数是几 答案:四十三 9.1根2米长的绳子将1只小狗拴在树干上,小狗虽贪婪地看着地上离它2.1米远的l根骨 头,却够不着,请问,小狗该用什么方法来抓骨头呢 答案:转过身用后腿抓 10.烟鬼甲每天抽50支烟,烟鬼乙每天抽10支烟。5年后,烟鬼乙抽的烟比烟鬼甲抽的还 多,为什么 答案:烟鬼甲抽得太多了早死了 11.一个数若去掉前面的第一个数字是11,去掉最后一个数字为50,原数是多少 答案:五十一 12.有一种细菌,经过1分钟,分裂成2个,再过1分钟,又发生分裂,变成4个。这样, 把一个细菌放在瓶子里到充满为止,用了1个小时。如果一开始时,将2个这种细菌放 入瓶子里,那么,到充满瓶子需要多长时间 答案:59分钟

(完整版)人教版小学三年级下册数学应用题(汇总)

人教版三年级下册应用题大集合服装店过去5天制作75套服装,现在每天制作18套,现在比过去每天多制作多少套服装? 一筐梨重25千克,一筐苹果比梨轻5千克,一筐香蕉比苹果重10千克,一筐香蕉重多少千克? 幼儿园买了15盒铅笔,买的蜡笔比铅笔多3盒,买的水彩笔是蜡笔的2倍。买了水彩笔多少盒? 妈妈今年32岁,比小玲大24岁,奶奶72岁。奶奶的年龄是小玲的几倍? 一个灯箱广告牌的长是80厘米,宽是40厘米,它的面积是多少平方厘米?合多少平方分米? 一张边长是30厘米的正方形纸,面积是多少平方厘米?合多少平方分米? 用一张长6厘米、宽4厘米的长方形纸剪一个最大的正方形。这个正方形的面积是多少平方厘米? 有两个完全一样的长方形,长都是4厘米,宽都是2厘米。用这两个长方形可以拼成一个正方形,拼成的正方形的面积和周长各是多少? 有一个长方形花圃,长18米,宽8米。(1)这个花圃的面积是多少平方米? (2)在花圃的四周围上篱笆,篱笆长多少米? 学校图书室要添置一批新书。《科学与发现》每套2本,每本4元。《数学故事》每本9元。《童话故事》每本38元。(1)用279元可以买多少本《数学故事》?(2)用450元买12本《童话故事》够不够?(3)用160元可以买多少套《科学与发现》? 每根不锈钢管8元8角,每根塑料管1元2角。一根不锈钢管长1.2米,一根塑料管长0.8米。(1)用小数表示两种水管每根的价钱。(2)一根不锈钢管比一根塑料管长多少米?一个正方形棋盘的边长是60厘米,要在棋盘上压一块与棋盘同样大的玻璃,这块玻璃的面积是多少平方厘米?合多少平方分米?

用一根24厘米长的铁丝围出边长是整厘米数的长方形或正方形,可以怎样围?有不同的围法吗? 有载重2吨、5吨、9吨的卡车各一辆。共有180吨货物。每种卡车各需要运多少次才能全部运完? 玫瑰花每束4枝,每枝2元。百合花每束9元。郁金香每束18元。(1)鲜花商店一天卖百合花的收入一共是810元。这一天共卖出百合花多少束?(2)用160元可以买多少束玫瑰花?(3)用500元买30束郁金香够不够? 一台磨面机每小时磨面粉6千克,要磨面粉126千克,需要几小时? 学校买来2000本练习本,分给16个班,每班120本,还剩多少本? 小刚家今年1-3月份共用电990 千瓦/时,小刚家平均每天用电多少千瓦/时? 手工课上,老师计划把同学们分成8个小组,给每个小组发3张彩色纸,实际上同学们是分成了6个小组活动的,那么每个小组可以得到几张彩色纸? 三年级同学收集了184千克废报纸。用32千克废报纸可以换4棵树苗,他们收集的废报纸可以换多少棵树苗? 商店第一天运来8箱饼干,每箱5千克,第二天运来150千克饼干,两天一共运来多少千克饼干? 同学们浇树,每人浇3棵,24人浇完。,几人浇完?(先补充条件再解答) (1)百货商店卖出3箱背心,每箱20件,每件24元,一共卖了多少元? (2)百货商店卖出3箱背心,每箱20件,一共卖了1400元,每件多少元? 1、有47千克苹果,平均分给幼儿园3个班。每班分得多少千克,还剩多少千克? 2、同学们去划船,男同学去了37人,女同学去了34人,每6人坐一条船。一共 需租多少条船? 3、王老师用100元买了3个排球,还找回4元。每个排球多少元?

小学数学竞赛试题大全

小学数学竞赛试题大全 1、一个正方形的边长增加3厘米,面积就增加39厘米,原来正方形的面积是多少平方厘米? 2.已知A、B两个数的最小公倍数是1000;A、C两数的最小公倍数和B、C两个数的最小公倍数都是2000;满足这个要求的数C有四个,分别是()、()、()、()。 3.已知1×2×3×4×5×6×……×n的末尾有连续100个0 ,那么n最小是多少? 4.有一列数:1、2、3、2、1、2、3、4、3、2、3、4、5、4、3、4、 5、6、5、4、5、……这列数中前240个数的和是()

5.把二进制数101011100写成十进制数是()把十进制数234写成二进制数是() 6.有9个连续的质数,它们的和偶数,则其中后5个数的平均数是( ) 7.数列1234,5678,9101112,……中,有一个十位数,这个十位数是() 8.个位是5的五位数中,能被9整除的所有的数的和是() 9.A是由2003个4组成的多位数,即4444……4。A是不是某个自然数B 的平方?如果是,写出B,如果不是,请说明理由

10.在一个正八边形的纸片内有100个点,以这100个点和八边形的8个顶点为顶点的三角形,最多能剪出多少个?最少可以剪多少个三角形? 11.分一堆苹果,每份3个,最后还剩一个;每份5个,最后还剩3个,每份7个最后还剩下5个,这堆苹果最少有多少个 12.从早晨7时到晚上7时,钟面上共有()次时针与分针成500角 13.从一块正方形木板上锯下5厘米宽的一个木条后,剩下的面积是750平方厘米。问锯下的木条的面积是多少平方厘米? 14.甲乙两人进行百米赛跑,当甲到达终点时,乙在甲后面20米,如果甲乙两人的速度保持不变,要使甲乙两人同时到达终点,甲的起跑线要比原来向后移动多少米?

小学数学三年级下册思维训练

三年级(下册)数学思维训练习题 单元目录 第一单元除数是一位数的除法 第二单元除数是一位数除法的应用题 第三单元年、月、日 第四单元年、月、日的应用题 第五单元平移和旋转 第六单元两位数乘两位数的乘法 第七单元两位数乘两位数的乘法应用题 第八单元认识千米 第九单元认识吨 第十单元轴对称图形 第十一单元认识分数(一) 第十二单元认识分数(二) 第十三单元长方形和正方形面积(一) 第十四单元长方形和正方形面积(二) 第十五单元统计与平均数 第十六单元认识小数(一) 第十七单元认识小数(二) 第十八单元观察物体

第一单元除数是一位数的除法 1、要使□36÷4的商是三位数,□里最小填()。 要使□36÷4的商是两位数,□里最大填()。 要使2□8÷8的商是三十多,□里可能填()。 2、一个三位数除以7商是75,有余数,余数最大是(),这时 被除数是()。 3、在□里填上什么数,商中间有0? 6)6□2 4、在□÷7=9……□中,被除数可能有几个?其中最大是几?最小是几? 5、 3 □ □)3 □□ □□ □□ □ 3 8 6、 7 □ 5)□□□ □ 5 □□ 4 5

第二单元除数是一位数除法的应用题 8、养殖场有300只鸡,鸡的只数是兔的3倍,把兔放在4个笼子里,平均每个笼子里有多少只兔? 9、两个水桶共盛水60千克,如果第一桶水倒出4千克则两个桶中的水同样多,求第一桶里原来盛水多少千克? 10、小明与小华共有图书160本,已知小明图书的本数是小华的3倍,求小明、小华各有图书多少本? 11、王庄有小麦、水稻田共180亩,小麦的亩数是水稻的2倍。王庄有小麦、水稻各多少亩? 12、学校图书馆有科技书和文艺书共2400本,文艺书的本数是科技书的4倍。两种书各有多少本? 13、爸爸与儿子的年龄和是45岁,又知爸爸的年龄是儿子的4倍,爸爸与儿子今年各多少岁? 第三单元年、月、日 14、从上午8时到下午5时经过()。

小学数学竞赛图解法(答案)

图解法(答案) 1.(24+1)÷(4+1)=5(条)。 答:电船有5条。 2.8张办公桌应付多少元? 68.50×8=548(元)。 每把椅子价是多少元? 68.50-36.50=32(元)。 买椅子共付多少元? 708-548=160(元)。 买了几把椅子? 160÷32=5(把)。 综合式: (708-68.50×8)÷(68.50-36.50)=5(把)答:买了椅子5把。 3.客车开出3.6小时行了多少公里? 50×3.6=180(公里)。 客车与货车共同行了多少公里? 488-180=308(公里)。 客车和货车每小时共行多少公里? 308÷2.8=110(公里)。 货车每小时行多少公里? 110-50=60(公里)。 综合式: (488-50×3.6)÷2.8-50=60(公里)

答:货车每小时行60公里。 4.第一层现有书多少本? 第一层原有书多少本? 42+10-16=36(本)。 第二层现有书多少本? 第二层原有书多少本? 63+10=73(本)。 第三层现有书多少本? 第三层原有书多少本? 84-20+16=80(本) 答:第一层原有书36本;第二层原有书73本;第三层原有书80本。5.张老师骑自行车25分钟行了多少公里? 李老师步行25分钟行了多少公里? 全程是多少公里?

综合式: 答:全程是4公里。 答:这桶煤油原有18公斤。 7.第二根电线的长是多少米? (14.8+1+2)÷(1+2+2)=3.56(米) 第一根电线的长是多少米? 3.56×2-1=6.12(米)。 第三根电线的长是多少米? 3.56×2-2=5.12(米) 答:三根电线分别为:6.12米、3.56米、5.12米。8.甲再从A地出发时,乙行了多少小时? 1.5×2+0.5=3.5(小时)。 甲再从A地出发时,甲乙两人相距多少里? 84-6×3.5=63(里)。 甲再从A地出发时,几小时与乙相遇? 63÷(8+6)=4.5(小时)。 综合式:[84-6×(1.5×2+0.5)]÷(8+6) =4.5(小时)。 答:甲再从A地出发,4.5小时与乙相遇。 6×(3.5+4.5)=48(里)

新人教版小学三年级数学下册公式概念(已整理)

新人教版小学三年级下册数学概念 班别: 姓名: 第一单元位置与方向 1、口诀要牢记:上北下南,左西右东。 2、东与西相对,南与北相对。(东北对西南,东南对西北) 东→南→西→北,就是按顺时针方向转。 3、地图通常就是按上北下南,左西右东绘制的。 做题时先标出上北下南,左西右东。 4、知道其中一个方向,可以通过顺时针方向按东、南、西、北的顺序确定其它的方向。 5、判断一个地方在什么方向,先要找到一个物体为观察点,再进行判断。 6、判断方向我们一般使用:指南针与借助身边的事物。我国早在两千多年就发明了指示方向的——司南。指南针就是用来指示方向的,它的一个指针永远指向(南方),另一端永远指向(北方) 7、会瞧简单的路线图,会描述行走路线 一定写清楚从哪儿向哪个方向走,走了多少米,到哪儿再向哪个方向走;同一个地点可以有不同的描述位置方式;(例如:学校在剧场的西面,在图书馆的东面,在书店的南面,在邮局的北面。)同一个地点有不同的行走路线;一般找比较近的路线走; 8、生活中的方位知识 (1)、早上起来,面向太阳,前面就是东,后面就是西,左面就是北,右面就是南。; 傍晚,面向太阳,前面就是西,后面就是东,左面就是南,右面就是北。晚上,面向北极星,前面就是北,后面就是南,左面就是西,右面就是东。 (2)、、我们通常用指南针来指示方向,指南针就是我国四大发明之一。生活中白天用太阳辨别方向,夜晚用北极星辨别方向。春天来了, 燕子从南方飞往北方,秋天来了,大雁从北方飞往南方过冬。 (3)、风从那边刮过来,那边就就是风向。影子的方向与太阳的方向相反;风筝,旗帜飘扬的方向与风向相反。 (4)、北斗星永远在北方。早上太阳在东方,中午在南方,傍晚在西方。 第二单元除数就是一位数的除法 1、口算时要注意: (1)0除以任何数(0除外)都等于0; (2)0乘任何数都得0; (3)0加任何数都得原数; (4)任何数减0都得原数。 2、整十、整百数除以一位数的口算:先用0前面的数除以一位数,再瞧被除数的末尾有几个0,就在得数的末尾加上几个0、 3、没有余数的除法: 被除数÷除数=商 商×除数=被除数 被除数÷商=除数 有余数的除法: 被除数÷除数=商……余数 商×除数+余数=被除数 (被除数—余数)÷商=除数 4、笔算除法顺序:确定商的位数,试商,检查,验算。余数的 (1)一位数除两位数(商就是两位数)的笔算方法:先用一位数除十位上的数,如果有余数,要把余数与个位上的数合起来,再用除数去除。除到被除数的哪一位,就把商写在那一位上面。 (2)一位数除三位数的笔算方法:先从被除数的最高位除起,如果最

小学数学竞赛一几种解题方法

一几种解题方法 1.28分。提示:按从多到少顺序枚举。如果小军是两个1角硬币,那么小红的三枚硬币不可能是18分;当小军是一个1角一个5分时,小红是一个1角,一个2分,一个1分。 2.5种。 3.495。解:因为93>700,所以只有下面三种可能: 13+33+53=153 13+33+73=371, 33+53+73=495,其中只有495是11的倍数。 4.286。解:此数是13的偶数倍,必能被26整除。由260依次往小试验,260-26=234,234-26=208,都不符合题意。再由260往大试验,260+26=286符合题意。 5.15。解:1与不小于4的任何自然数都不满足题意,所以四个数中没有1。取2,3,4,a,前三个数满足条件,a=5不满足条件,a=6满足条件。所求数为2+3+4+6=15。 6.8种。解:将四个瓶子依次记为A,B,C,D,将四张标签依次记为a,b,c,d。假设A贴对了,其余的都贴错了,有两种情况: ①Aa,Bc,Cd,Db;②Aa,Bd,Cb,Dc。 同理B,C,D贴对了,其余的都贴错了,也各有两种情况。共8种。 7.10种。提示:有0,0,3;0,1,2;0,2,1;0,3,0;1,0,2;1,1,1;1,2,0;2,0,1;2,1,0;3,0,0十种方法。 8.7。解:不拆盒可买的节数有3,5,8,9,10,…因为超过10的数都可以由8,9,10中的某个数加3的倍数形成,而8,9,10都可以不拆盒,所以买7节以上(不含7)都不必拆盒。 9.11。提示:与第8题类似。 10.18支、10支、6支、4支。提示:因为总的铅笔数不多,故可依次假设丁有2支、3支、4支……铅笔。 11.21个。 提示:乙的红球、白球都是偶数。因为甲的红球数是乙的白球数的2倍,并且不超过10,所以乙的白球数只能是2或4。

小学数学趣味题

趣味数学题库 姐俩看电影 小芳、小花姐妹二人从家里出发到电影院看电影,小芳每小时走5公里,小花每小时走3公里,她们同时出发1小时后,姐姐又回家拿东西再去追妹妹,妹妹仍以原速前进,最后二人同时到达电影院。求从家里到电影院之间的距离? 小马虎数鸡 春节里,养鸡专业户小马虎站在院子里,数了一遍鸡的总数,决定留下1/2外,把1/4慰问解放军,1/3送给养老院。他把鸡送走后,听到房内有鸡叫,才知道少数了10只鸡。于是把房内房外的鸡重数一遍,没有错,不多不少,正是留下1/2的数。小马虎奇怪了。问题出在哪里呢?你知道小马虎在院里数的鸡是多少只吗? 来了多少客人 一天,小林正在家里洗碗,小强看见了问道:“怎么洗那么多的碗?”“家里来了客人了。”“来了多少人?”小林说:“我没有数,只知道他们每人用一个饭碗,,二人合用一个汤碗,三人合用一个菜碗,四人合用一个大酒碗,一共用了15个碗。”你知道来了多少客人吗? 称珠子 有243颗外形一模一样的珠子,其中有一颗稍重一点。用一架没有砝码的天平,至少称几次才能找出这颗珠子来? 分梨 箱子里放着一箱梨,第一个人拿了梨总数的一半又多半只,第二个人拿了剩下梨的一半又多半只,第三个人拿了第二次剩下的一半又多半只,第四个人3拿了第三次剩下的一半又多半只,第五个人拿了第四次剩下的一半又多半只。这时箱子里的梨正好拿完,而且每人手里的梨都没有半只的,请问箱子里原来有多少只梨? 如何分组 暑假里,班里要作社会调查,要分成15个小组,班里有赵、钱、孙、李、周各6位同学,要使每个小组的姓都不同,该如何分呢? 巧算星期 今年的十月一日是星期一,明年的十月一日是星期几?请写出简便算法来? 谁跑得快 小伟与小林百米赛跑,结果当小伟跑到终点时,小林只跑了95米。小林要求再跑一次,这次小伟的起跑线比小林退后5米,如果他们都用原来的速度跑,那么同时到达终点吗? 火车过桥 南京长江大桥的铁路桥共长6772米,一列货车长428米,每秒行驶20米,请问全车通过大桥要多少时间? 开锁问题 用外观一模一样的钥匙试开10把锁,最多试多少次,就可以分辨出哪把钥匙配哪把锁的?

小学三年级下学期数学概念

小学三年级下学期数学概念 第一单元:位置与方向 1、地图通常是按“上北下南,左西右东”绘制的。 2、东与西相对,南与北相对 第二单元:除数是一位数的除法 1、笔算除法的要领:要从被除数的最高位除起,除到被除数的哪一位,商就写在哪一 位的上面,如果哪一位不够商1,就在那一位上面商0。 2、注意:每次除得的余数一定要比除数小。 3、验算的公式:被除数=商×除数+余数或者被除数=商×除数 4、估算的方法:除数不变,把被除数估成最接近的整十,整百数,或依照乘法口诀来 估算。 5、判断商是几位数的方法:如果被除数的最高位小于除数时,商就是两位数,如果被 除数的最高位大于或等于除数时,商就是三位数。 6、0除以任何不是0的数都得0;任何数除以1或乘1都得原数。 第三单元:统计 复式统计表可以清晰的反映数据的情况。 第四单元:两位数乘两位数 笔算方法:相同数位对齐,用第二个因数个位的数去乘第一个因数每一位上的数,从个位乘起,乘得的积表示多少个一,再用第二个因数十位上的数去乘第一个因数,也是从个位乘起,乘得的积表示多少个十。 第五单元:面积 1、物体表面的大小,就是它们的面积。比较两个图形面积的大小,要用统一的面积单位来 测量。 2、常用的面积单位有平方米、平方分米、平方厘米,用字母表示是m2、dm2、cm2,相邻 两个面积单位的进率是100。 3、长度单位有千米、米、分米、厘米、毫米,相邻两个单位之间的进率一般是10,特殊的 有1千米=1000米。 4、质量单位有吨、千克、克,相邻两个单位之间的进率是1000。 1吨=1000千克1千克=1000克 5、时间单位有时、分、秒,时间单位的进率是60. 1时=60分1分=60秒 6、周长在四周,面积在其中。周长一条线,面积一大片。周长与面积的区别:概念不相同,单位不相同,计算方法不相同。 7、长方形的周长=(长+宽)×2 长方形的面积=长×宽 正方形的周长=边长×4 正方形的面积=边长×边长 8、边长是1厘米的正方形,面积是1平方厘米;边长是1分米的正方形,面积是1平方分米;边长是1米的正方形,面积是1平方米。 9、边长是4米的正方形,它的周长和面积不相等,因为单位不同不能比较。

小学高段趣味数学题附答案完整版

小学高段趣味数学题附 答案 HEN system office room 【HEN16H-HENS2AHENS8Q8-HENH1688】

小学高年级数学趣味题 1.8个数字“8”,如何使它等于1000? 2. 答案:8+8+8+88+888 3.小强数学只差6分就及格,小明数学也只差6分就及格了,但小明和小强的分数 不一样,为什么? 答案:一个是54分,一个是0分 4.一口井7米深,有只蜗牛从井底往上爬,白天爬3米,晚上往下坠2米。问蜗牛 几天能从井里爬出来? 5. 答案:5天 6.某人花19快钱买了个玩具,20快钱卖出去。他觉得不划算,又花21快钱买进, 22快钱卖出去。请问它赚了多少钱? 答案:2元 7.100个包子,100个人吃,1个大人吃3个,3个小孩吃1个,多少个大人和多少 小孩刚好能吃完? 答案:25个大人,75个小孩 8.小王去网吧开会员卡,开卡要20元,小王没找到零钱,就给了网管一张50的, 网管找回30元给小王后,小王找到20元零的,给网管20元后,网管把先前的50 元还给了他,请问谁亏了? 答案:网管亏了30元 9.每隔1分钟放1炮,10分钟共放多少炮? 答案:11炮 10.一个数去掉首位是13,去掉末位是40.请问这个数是几? 答案:四十三 11.1根2米长的绳子将1只小狗拴在树干上,小狗虽贪婪地看着地上离它2.1米 远的l根骨头,却够不着,请问,小狗该用什么方法来抓骨头呢? 12. 答案:转过身用后腿抓 13.烟鬼甲每天抽50支烟,烟鬼乙每天抽10支烟。5年后,烟鬼乙抽的烟比烟鬼 甲抽的还多,为什么? 答案:烟鬼甲抽得太多了早死了 14.一个数若去掉前面的第一个数字是11,去掉最后一个数字为50,原数是多 少? 答案:五十一

小学三年级下册数学各单元练习题及答案

小学三年级下册数学各单元练习题 位置与方向 一、填一填 1、小东早晨上学,他面向太阳,他的前面是(东),后面是(西),左面是(北),右面是(南)。2 (1)小刚出门向(北)走(90)米就到学校了。 (2)小红出门向(北)走(60)米,再向(东)走(100)米,最后向(北)走(50)米就到学校了。 (3)小明出门后向(西)走(70)米,再向(北)走(50)米就到学校了。 3、帮小动物找家 小鸡的家 小狗的家小鸭的家小猪的家 小猫的家 (1)小鸡的家在小鸭的家的(北)面,小猪的家在小鸭的家的(东)面。 (2)小猫的家在小鸭的家的(南)面,小猪的家在小鸡的家的(东南)面。 二、选择题

1、早上太阳从(A)方升起,傍晚在( C )方落下。 A.东 B、西 C、南 D、北 2、小红向东走,迎面走来小明,小明向(B)面走。 A.东 B、西 C、南 D、北 3、大树的影子在东边,那么太阳在大树的(B )面。 A.东 B、西 C、南 D、北 4、冬冬座位的西北方向是李伟的座位,那么冬冬的座位在李伟的座位的(B)方向。 A.西北 B、东南 C、东北 D、西南 5、西南与(A)相对。w W w . X k b 1.c A.东北 B、东南 C、西北 三、一起去逛街 我从新华路乘公交车向(东北)方向坐(2 )站地到公园,再向(东南)方向坐(2)站地到书店,最后向(东)方向坐(3)站地就到火车站了。 口算除法练习题 1、口算下面各题. 200÷5 =40 120÷6 =20 2400÷6= 40 100÷5=20 3×200=600 350÷7=50 500÷5=100 100×6=600 2000÷2=1000 1200÷4=300 700÷7=100 2000÷5=400 160÷4=40 120÷3=40 1500÷3=500 800÷8=100 630÷9=70 400÷4=100 4900÷7=700 540÷6=90

小学数学竞赛四 其它类型的问题

四其它类型的问题 例1如图7-7,一支箭头表示一段有方向的路,试计算从A至I有多少条不同的路可走? 分析因为路数总是有限的,于是很自然的想法是用枚举法试试,但是试过几条路后,就会觉得由于道路的错综复杂,哪些路算过,哪些路还没有算过都搞不清了,所以我们不妨换一条思路,用倒推法试一试. 从图上可以看出来,要想到I,就必须经过F、M、H之一,因此从A 至F,M,H各有几条路.其总和便是从A到I所有不同的路数,对F,M,H又可做同样的倒推分析,这样一步一步地倒推下去,最后归结为找A至B,C,D的不同路数,因此再反推过去,便得到如下的计算方法:如果把A至某点的路数记在表示该点的字母旁边,那么,要求A到达某点的路数,只要看有几只箭头到达该点,这几只箭头的尾部数据之和即为所求.为统一起见,A点数据记为1,于是由A到B,C,D的路数分别为1,3,1,到E的路数为1+3+1=5,其它类推,最后得A至I的路数为24+12+6=42,见图7-8. 例2有一天,三位小朋友在图书馆相会,其中一个说:我每隔一天来一次.第二个说:我每隔两天来一次.第三个说:我每隔三天来一次.管理员告诉他们说:每逢星期三闭馆,小朋友们说:如果预定来的日子正好闭馆,那就次日来,从今天开始,他们按上述办法来,下一次在星期一他们三人又在图书馆相聚,上次谈话离这个星期一最近可能是星期几? 分析设这三个小朋友分别为A、B、C,对A来说,星期一的前一次是星期六,再前一次是星期四,再前一次是星期几呢?有两种可能性,星期二或星期一,对B,C二人也作类似的倒推分析,我们把分析的结果列成下表,表中的“√”表示可能去图书馆.

解从表中看出,对A和C来说,都在星期四去了图书馆,而星期四的前一次有两种可能性.所以离星期一最近的相遇时间可能是星期六. 例3 把123,124,125三个数分别写在图7-9所示的A,B,C三个小圆内,然后按下面的规则修改这三个数: 第一步:把B中数改为A中数与B中数之和; 第二步:把C中数改成B中(已改过)的数与C中数之和; 第三步:把A中数改为C中(已改过)的数与A中数之和. 再回到第一步,循环做下去. 如果在某一步做完之后,三个圆圈的数都变成了奇数,则停止运算. 为了尽可能多运算几步,那么124应填在哪一个圆圈里? 分析因为题目的要求,只是经过多少次运算后使圆圈中的数全部变为奇数,并不要求算出最后圆圈中的数,因此我们可以把123,124,125看成奇数,偶数,奇数,这个问题变成为一偶数两奇数分别放入三个圆圈中,按规则修改圆圈中的数,为使A,B,C中的数都变成奇数,偶数应放入哪个圆圈,就可以使运算步数尽可能多. 解为了叙述方便,我们用0表示偶数,用1表示奇数,三个圆圈分别用甲、乙、丙表示,从开始一步一步地倒推回去,如图7-10所示.

小学低年级趣味数学题及答案

低年级趣味数学题 1、填数10、7、4、() 2、5、()、11、14、 20、16、()、8、4 15、3、13、3、11、3、()、() 8,(),12,14,()(),11,9,7 0、3、()、9、12 ()、()、15、20、25 2、河里有一行鸭子,2只的前面有2只,2只的后面有2只,2 只的中间还有2只,共有几只鸭子? 3、哥哥给弟弟4支铅笔后,哥哥与弟弟的铅笔就一样多了,原来哥哥比弟弟多几支铅笔? 4、在一排10名男同学的队伍中,每两名男同学之间插进1名女同学,请你想一想,可以插进多少名女同学? 5、一杯牛奶,小明喝了半杯,又倒满了水,又喝了半杯后,再倒满水后,一饮而进,他喝了几杯水?几杯奶? 6、有9棵树,种成3行,每行4棵,应该怎样种?画出来。 7、有3只猫同时吃3只老鼠共用3分钟,那么100只猫同时吃100只老鼠,需要多少分钟? 8、把一根5米长的木头锯成5段,要锯多少次? 9、小朋友们排成一排,小华前面有4人,后面有10人,小华排在第几名?这一排一共有多少人? 10、甲、乙两个相邻的数的和是19,那么,甲数是多少?乙数是多少? 11、小明有10本书,小红有6本书,小明给小红多少本书后,两人

的书一样多? 12、小朋友们吃饭,每人一只饭碗,2人一只菜碗,3人一只汤碗,一共用了11个碗,算一算,一共有几人吃饭? 13、游乐场中,小红坐在环形的跑道上的一架游车上,他发现他前面有5架车,后面也有5架车,你认为包括小红坐的车,跑道上一共有多少架车? 14、爸爸买来两箱梨,第二箱比第一箱轻8千克,爸爸要从第几箱中搬出几千克到第几箱,两箱的梨就一样重了? 15、有一排花共13盆,再每两盆花之间摆1棵小树,一共摆了多少棵小树? 16、一根绳子对折、再对折后,从中间剪开,这根绳子被分成了几段? 17、科学家在实验室喂养一条虫子,这种虫子生长的速度很快,每天都长长1倍,20天就长到20厘米,问:当它长到5厘米时用了几天? 18、池塘里的睡莲的面积每天增长一倍,6天可长满整个池塘,需要几天睡莲长满半个池塘? 19、教室里有10台风扇全开着,关掉4台,教室里还有多少台风扇? 20、如果A+3=B+5,那么,A和B两个数谁大?大多少? 21、小朋友们站一排,从前往后数小红排第4名,从后往前数,小红也排第4名,这一排一共有多少人? 22、小朋友们站一排,小红前面有4个人,小红后面也有4个人,这一排一共有多少人? 23、小朋友们站一排,从前面数小红是第4名,她后面还有4个人,

最新人教版小学三年级数学下册知识点大全

第一单元位置与方向 1、①(东与西)相对,(南与北)相对, (东南—西北)相对,(西南—东北)相对。 ②清楚以谁为标准来判断位置。 ③理解位置是相对的,不是绝对的。 2、地图通常是按(上北、下南、左西、右东)来绘制的。 (做题时先标出北南西东。) 3、会看简单的路线图,会描述行走路线。 一定写清楚从哪儿向哪个方向走,走了多少米,到哪儿再向哪个方向走。同一个地点可以有不同的描述位置的方式。(例如:学校在剧场的西面,在图书馆的东面,在书店的南面,在邮局的北面。)同一个地点有不同的行走路线。一般找比较近的路线走。 4.、指南针是用来指示方向的,它的一个指针永远指向(南方),另一端永远指向(北方)。 5.、生活中的方位知识:

①北斗星永远在北方。 ②影子与太阳的方向相对。 ③早上太阳在东方,中午在南方,傍晚在西方。 ④风向与物体倾斜的方向相反。 (刮风时的树朝风向相对的方向弯,烟朝风向相对的方向飘……) 第二单元除数是一位数的除法 1、口算时要注意: (1)0除以任何数(0除外)都等于0; (2)0乘以任何数都得0; (3)0加任何数都得任何数本身; (4)任何数减0都得任何数本身。 2、没有余数的除法: 被除数÷除数=商 商×除数=被除数 被除数÷商=除数 有余数的除法: 被除数÷除数=商……余数 商×除数+余数=被除数 (被除数—余数)÷商=除数 3、笔算除法顺序:确定商的位数,试商,检查,验算。

(1)一位数除两位数(商是两位数)的笔算方法:先用一位数除十位上的数,如果有余数,要把余数和个位上的数合起来,再用除数去除。除到被除数的哪一位,就把商写在那一位上面。 (2)一位数除三位数的笔算方法:先从被除数的最高位除起,如果最高位不够商1,就看前两位,而除到被除数的哪一位,就要把商写在那一位上,假如不够商1,就在这一位商0;每次除得的余数都要比除数小,再把被除数上的数落下来和余数合起来,再继续除。 (3)除法的验算方法: 没有余数的除法的验算方法:商×除数:被除数; 有余数的除法的验算方法:商×除数+余数=被除数。 4、基本规律: (1)从高位除起,除到哪一位,就把商写在那一位; (2)三位数除以一位数时百位上够除,商就是三位数;百位上不够除,商就是两位数;(最高位不够除,就看两位上商。) (3)哪一位有余数,就和后面一位上的数合起来再除; (4)哪一位上不够商1,就添0占位;每一次除得的余数一定要比除数小。 第二单元课外知识拓展 5、2、3、5倍数的特点

重点小学新六年级数学奥赛竞赛题附参考答案

学习奥数的重要性 小学六年级数学奥赛竞赛题1.学习奥数是一种很好的思维训练。奥数包含了发散思维、收敛思维、换元思维、反向思维、逆向思维、逻辑思维、空间思维、立体思维等二十几种思维方式。通过学习奥数,可以帮助孩子开拓思路,提高思维能力,进而有效提高分析问题和解决问题的能力,与此同时,智商水平也会得以相应的提高。 2.学习奥数能提高逻辑思维能力。奥数是不同于且高于普通数学的数学内容,求解奥数题,大多没有现成的公式可套,但有规律可循,讲究的是个“巧”字;不经过分析判断、逻辑推理乃至“抽丝剥茧”,是完成不了奥数题的。所以,学习奥数对提高孩子的逻辑推理和抽象思维能力大有帮助 3.为中学学好数理化打下基础。等到孩子上了中学,课程难度加大,特别是数理化是三门很重要的课程。如果孩子在小学阶段通过学习奥数让他的思维能力得以提高,那么对他学好数理化帮助很大。小学奥数学得好的孩子对中学阶段那点数理化大都能轻 松对付。 4.学习奥数对孩子的意志品质是一种锻炼。大部分孩子刚学奥数时都是兴趣盎然、信心百倍,但随着课程的深入,难度也相应加大,这个时候是最能考验人的:少部分孩子凭着天分,凭着在困难面前的百折不挠和愈挫愈坚的毅力,坚持了下来、学了进去、收到了成效;一部分孩子在家长的“威逼利诱”之下,硬着头皮熬了下来;不少孩子更是或因天资不足、或惧怕困难、或受不了这份苦、再或是其它原因而在中途打了退堂鼓。我以为,只要能坚持学下来,不论最后取得什么样的结果,都会有所收获的,特别是对孩子的意志力是一次很好的锻炼,这对他今后的学习和生活都大有益处。 小学六年级数学奥赛竞赛题 一、计算 1.×+÷+×. 2.×+×. 3.1999+999×999. 4.8+98+998+9998+99998. 5.(﹣×25十75%×)÷15×1997. 二、填空题 6.六(1)班男、女生人数的比是8:7. (1)女生人数是男生人数的_________(2)男生人数占全班人数的_________ (3)女生人数占全班人数的_________(4)全班有45人,男生有_________人. 7.甲数和乙数的比是2:5,乙数和丙数的比是4:7,已知甲数是16,求甲、乙、丙三个数的和是_________.8.甲数和乙数的比7:3,乙数和丙数的比是6:5,丙数是甲数的_________,甲数和丙数的比是_________:_________. 9.的倒数是_________,的倒数是_________. 10.一根铁丝长3米,剪去1/3后还剩_________米;一根铁丝长3米,剪去1/3米后还剩_________米.11.甲、乙合做一件工作,甲做的部分占乙的,乙做的占全部工作的_________. 12.周长相等的正方形和圆形,_________的面积大. 13._________÷40=15:_________═=_________% 14.把、、37%、按从大到小的顺序排列是_________. 15.4米是5米的_________%,5米比4米多_________%,4米比5米少_________%

小学六年级趣味数学

1.百个和尚百个粑,大和尚每人粑四个,小和尚四人一个粑,大小和尚各有几? 2.小鸡、小狗七十九,两百只脚地上走,想一想,算一算,多少只鸡?多少只狗? 3.妈妈给小明一个大盒子,里面装着6个纸盒子,每个纸盒子又装4个小盒子,小明一共有几个盒子? 4.王老太去集市卖鸡蛋,第一个人买走篮子里鸡蛋的一半又一个,第二人买走剩下鸡蛋的一半又一个,这时篮子里还剩一个鸡蛋,王老太共卖出几个鸡蛋? 5.猜谜语: 5、4、3、2、1(打一数学名词)看谁力量大(打一数学名词) 考试作弊(打一数学名词)1×1(猜一成语)3.4(猜一成语)30天÷2(打一汉字) 老爷爷参加赛跑(打数学家名)72小时(打一汉字) 6.按规律填空:1,1,2,3,5,8,13,21,(),() 1,5,9,13,(),21,() 1,3,9,27,(),243,() 1,2,4,7,(),16,() 1,3,7,15,(),63,() 7.井深8米,一只青蛙从井底往上跳,每次跳3米,又滑下2米,那么它要跳几次才能到达井口? 8.大小两只青蛙比赛捉虫子,大青蛙比小青蛙捉得多。如果小青蛙把捉的虫子给大青蛙3只,则大青蛙捉的就是小青蛙的3倍。如果大青蛙把捉的虫子给小青蛙15只,则大小青蛙捉的虫子一样多。你知道大小青蛙各捉了多少只虫子吗? 9.小猴子从300米远的地方往回抬一个大西瓜,需要2个小猴子一起抬,现在由3个小猴子轮流参加抬,请你算一下,每个小猴子抬西瓜平均走了多少米 10.一只笼子里有白兔、黑兔若干只,如果拿出2只黑兔,白兔黑兔只数相等,如果拿出1 只白兔,黑兔只数是白兔的2倍。问白兔、黑兔各多少只 11.太阳落下西山坡,鸭儿嘎嘎要进窝。四分之一岸前走,一半的一半随水波;身后还跟八 只鸭,我家鸭子共几多? 12.有一人老婆怀孕了,他在临死前立了个遗嘱,如果生了男孩,他的遗产2/3分配给儿子, 1/3分配给老婆;如果生了女孩,1/3分给女儿,2/3分给老婆。结果他老婆生了龙凤胎,请问,这时候遗产应该怎么分配。 13.一桶水可以装满10碗或12杯,倒入5杯水和3碗水在空桶内,水面高度占桶高度的多少? 14.电影票480张,如果先给五年级学生,剩下的只能给六年级一半的学生,如果下面分给 六年级,剩下的给五年级,剩下的给五年级,那么五年级会有1/3的学生分不到票。五、六年级各有学生多少人? 15.桌子的单价是椅子的4倍,3套桌椅360元,每张桌子的单价是多少元? 16.父亲的年龄是小聪年龄的9倍,母亲的年龄是小聪年龄的7.5倍,父亲比母亲大6岁, 小聪今年多少岁? 17.一次数学知识抢答比赛中,共10道题,规定答对一题得5分,答错一题倒扣3分,强强 得了34分,他做对了几道题?

(完整版)小学三年级下册数学试题三篇

小学三年级下册数学试题三篇 【篇一】 一、选择。 1.太阳( )是东升西落。 A.一定 B.不一定 C.不会 2.与北极星相对的方向是( ) 。 A.东 B.南 C.西 3.小明座位的西南方向是张强的座位,那么小明在张强的( )方向。 A.东南 B.西北 C.东北 4.三(1)班教室的黑板在教室的西面,那么老师讲.课时面向( )面。 A.东 B.南 C.西 D.北 5.张丽面向南站立,当她向后转之后,她的左面是( ),右面是( )。 A.东 B.西 C.北 二、填空。 1.把手表平放在桌面上,用数字12 正对着北方。正对着南方的是数字( );数字3 正对着( )方。 2.小铃面向西站立,向右转动两周半,面向( );向左转动l周半,面向( )。 三、算一算,分分类。 (1)把得数小于50的写在西面。 (2)把得数在50~100的写在东面。 (3)把得数在100—200的写在北面。 (4)把得数在200以上的写在南面。 四、判断,对的画“√”,错的画“×”。 1.人的影子在西方,太阳应在东方。( ) 2.和西北相对的方向是西南。( ) 3.在森林中可以利用树叶的疏密来识别方向。( ) 4.面对早晨的太阳,你的右手边是南方。( ) 五、应用题。 1.小强的家门面向东,放学回家后站在门前,面向家门,他的前后左右分别是什么向? 2.小明和小立背对背站立,小明向北走150米,小立向南走120米,两人相距多远? 3.小娟向东走5步,然后向西走4步,再向东走3步,再向西走2步,再向东走1步,现在小娟在出发点的什么方向几步的地方? 4.(探究题)小明面向东向前走5步,左转向前走4步,再左转向前走5步,现在小明面向什么方向?如果想尽快回到原地,可以怎样走? 5.(情景题)夺红旗,争第一。 小熊从A点出发,—共有几条通往小红旗的路线?哪条路线是最近的? 描述出你认为最近的走法。 6.(开放题)李芳与张林相邻,李芳东面有25名同学,张林西面有5名同学,这一排中一共有多少名同学?

相关文档
最新文档