长沙理工大学数据结构栈的实现及应用算术表达式求值实验报告

实验报告

年级班号学号姓名

实验名称：栈的实现及其应用：算术表达式的计算实验日期2016年12月2日

计算机科学与技术系

2016年制

一、实验环境

32位操作系统下的Window平台Microsoft Visual C++

二、实验目的

掌握栈的实现及使用

三、实验内容

1.实现栈的存储结构

2.实现栈的基本操作的有关算法

3.利用栈解决*算术表达式求值问题

四、数据结构与算法思想描述

顺序读取中缀表达式：

1、当遇到数字时，将数字入数字栈

2、当遇到操作符时，与操作符栈栈顶比较：

If(当前操作符优先级大于操作符栈栈顶的优先级)

If(非”)”操作符)

将当前操作符进操作符栈;

Else

While(操作符栈栈顶不等于”(“)

取操作符栈栈顶及数字栈的两个数进行运算，并将结果压入数字栈;

Else

If(非(“操作符)

While(操作符栈栈顶不等于”(“)

取操作符栈栈顶及数字栈的两个数进行运算，并将结果压入数字栈;

Continue;(直到当前操作符比栈顶操作符优先级大)

Else

将当前操作符进操作符栈;

3、While(操作符栈非空)

操作符栈栈顶及数字栈的两个数进行运算，并将结果压入数字栈;

4、在数字栈取最后结果并输出。

五、程序清单

//10*8^2+16.3+5*(5.2*5+3.01)/4-(-10)+0.1000060+4.00416-40 = 666.666666 //100+(-100)-(-10^2) = 100

//(((2016-2017+(((2015-2014)))))) = 0

//-1+(((((((((1^0))))))))+100%10^2 = 0

#include

#include

#include

#include

#include

#include

using namespace std;

const int MAX = 105;

typedef double Type;

typedef struct

{

Type TypeStack[MAX];

char charStack[MAX];

int TypeTop, charTop;

}Stack;

//初始化栈

void InitStack(Stack *S)

{

S->charTop = S->TypeTop = 0; }

//判断charStack是否为空

bool IsEmpty_Char(Stack S)

{

return S.charTop == 0;

}

//判断TypeStack是否为空

bool IsEmpty_Type(Stack S)

{

return S.TypeTop == 0;

}

//判断charStack是否为满

bool IsFull_Char(Stack S)

{

return S.charTop == MAX;

}

//判断TypeStack是否为满

bool IsFull_Type(Stack S)

{

return S.TypeTop == MAX;

}

void Push_Char(Stack *S, char ch)

{

//charStack不为满则入栈，否则输出提示

if(!IsFull_Char(*S))

S->charStack[S->charTop++] = ch;

else

cout << " The CharStack Is Full! " << endl; }

void Push_Type(Stack *S, Type a)

{

//TypeStack不为满则入栈，否则输出提示

if(!IsFull_Type(*S))

S->TypeStack[S->TypeTop++] = a;

else

cout << " The TypeStack Is Full! " << endl; }

char Pop_Char(Stack *S)

if(!IsEmpty_Char(*S))

{

S->charTop--;

return S->charStack[S->charTop];

}

else

cout << " The CharStack Is Empty! " << endl;

return -1;

}

Type Pop_Type(Stack *S)

{

if(!IsEmpty_Type(*S))

{

S->TypeTop--;

return S->TypeStack[S->TypeTop];

}

else

cout << " The TypeStack Is Empty! " << endl;

return -1;

}

char Top_Char(Stack S)

{

if(!IsEmpty_Char(S))

return S.charStack[--S.charTop];

else

cout << " The CharStack Is Empty! " << endl;

return -1;

}

Type Top_Type(Stack S)

{

if(!IsEmpty_Type(S))

return S.TypeStack[--S.TypeTop];

else

cout << " The TypeStack Is Empty! " << endl;

return -1;

}

Type Calculate(Type left, Type right, char op)

{

Type value = 0;

switch(op)

{

case '+': value = left + right; break;

case '-': value = left - right; break;

case '*': value = left * right; break;

case '/': if(right != 0)

value = left / right;

else

cout << "被除数不能为零!" << endl;

break;

case '%': i f(right != 0)

value = (int)left % (int)right;

else

cout << "被余数不能为零!" << endl;

break;

case '^': value = pow(left,right);

/*value = 1;

if(right >= 0)

while(right--)

value *= left;

else

{

right = -right;

while(right--)

value /= left;

}*/

}

return value;

}

void Computer(char *mid_equotion, Type len)

{

Type right, left , result;

char *p_mid_equotion = mid_equotion;

char after_equotion = ' ';

map Oper;

Oper['#'] = 1; Oper['('] = 2; O per['+'] = 3;

Oper['-'] = 3; O per['*'] = 4; Oper['/'] = 4;

Oper['%'] = 4; Oper['^'] = 5; Oper[')'] = 6;

Stack MyStack;

InitStack(&MyStack);

Push_Char(&MyStack,'#');

char top_oper, current_oper;

for(;*p_mid_equotion != '\0';)

{

top_oper = Top_Char(MyStack);

current_oper = *p_mid_equotion;

if(!Oper[current_oper])

{

Push_Type(&MyStack,strtod(p_mid_equotion, &p_mid_equotion));

continue;

}//end if

else //为操作符

{

if(Oper[current_oper] > Oper[top_oper])

{

if(current_oper != ')')

Push_Char(&MyStack,current_oper);

else

{

while(top_oper != '(')

{

right = Pop_Type(&MyStack);

if(!IsEmpty_Type(MyStack))

left = Pop_Type(&MyStack);

else

left = 0;

Push_Type(&MyStack,Calculate(left, right, Top_Char(MyStack)));

Pop_Char(&MyStack);

top_oper = Top_Char(MyStack);

}

Pop_Char(&MyStack);

}//end else

}//end if

else

{

if(current_oper == '(')

{

Push_Char(&MyStack,current_oper);

if(*(p_mid_equotion + 1) == '-')

Push_Type(&MyStack,0);

}

else

{

right = Pop_Type(&MyStack);

if(!IsEmpty_Type(MyStack))

left = Pop_Type(&MyStack);

else

left = 0;

Push_Type(&MyStack,Calculate(left, right, top_oper));

Pop_Char(&MyStack);

continue;

}

}//end else

}//end else

p_mid_equotion++;

}//end for

top_oper = Pop_Char(&MyStack);

while(top_oper != '#')

{

right = Pop_Type(&MyStack);

if(!IsEmpty_Type(MyStack))

left = Pop_Type(&MyStack);

else

left = 0;

Push_Type(&MyStack,Calculate(left, right, top_oper));

top_oper = Pop_Char(&MyStack);

}

// cout << setprecision(6) << "\nThe Result = " << (result = Pop_Type(&MyStack)) << endl;

printf("The Result = %lf\n\n",(result = Pop_Type(&MyStack)));

}

int main()

{

char s[MAX] = "";

Type i = 0;

cout << "请输入你要求值的表达式!(以-1结束)\n";

while(cin >> s && strcmp(s,"-1") != 0)

{

Computer(s,strlen(s));

cout << "请输入你要求值的表达式!(以-1结束)\n";

}

return 0;

}

六、程序执行结果及其分析

对“+” , “-” , “*” , “/” , “%” , “^”运算的实现

可运算多位数和小数，求余，求平方，括号里包含负数如（-1），及首个数字为负数如-1+1