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C语言第十章习题答案

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第10章C语言程序设计习题答案

第10章C语言程序设计习题答案

C 语言程序设计( Visual C++6.0 环境)》习题答案习题十、思考题1．简述公有类型成员与私有类型成员的区别。

公有(public) 类型成员不但可以被类的成员函数访问，而且可以被外界访问，所以说公有类型定义了类的外部接口。

私有(private) 类型成员只能被类的成员函数访问，外界不能直接访问它。

类的数据成员一般都应该声明为私有成员。

2．简述构造函数与析构函数的作用。

构造函数的作用就是在对象在被创建时利用特定的值构造对象，将对象初始化。

析构函数的作用与构造函数正好相反，它是用来在对象被删除前进行一些清理工作。

析构函数调用之后，对象被撤消了，相应的内存空间也将被释放。

3．简述什么是友元函数。

友元函数是在类定义中由关键字friend 修饰的非成员函数。

友元函数可以是一个普通函数，也可以其它类中的一个成员函数，它不是本类的成员函数，但它可以访问本类的私有成员和保护成员。

4．简述公有继承、私有继承和保护继承三种继承方式的区别。

⑴、当类的继承方式为公有(public 继承)时，基类的公有(public )成员和保护( protected )成员仍然成为派生类的公有成员和保护成员，而基类的私有成员不能被派生类访问。

⑵、当类的继承方式为保护( protected )继承时，基类的公有(public )成员和保护( protected )成员将成为派生类的保护成员，而基类的私有成员不能被派生类访问。

⑶、当类的继承方式为私有(private )继承时，基类的公有(public )成员和保护(protected )成员将成为派生类的私有成员，而基类的私有成员不能被派生类访问。

5．定义一个圆柱体类，其属性为圆柱体的底面半径和高，能计算出圆柱体的体积。

#include<iostream.h>class cylinder{public:cylinder(float r,float h){radius=r;height=h;}float Volume();private:float radius;float height;};float cylinder::Volume(){return 3.14*radius*radius*height;}void main(){float r,h;cout<<" 请输入圆柱体的底面半径和高:"; cin>>r>>h;cylinder x(r,h);cout<<x.Volume()<<endl;}6．从第 5 题中定义的圆柱体类中派生出圆锥类，覆盖计算体积的成员函数。

《C语言程序设计》教材习题答案第10章

《C语言程序设计》教材习题答案第10章

一、选择题1. 设已定义“int a，* p”，下列赋值表达式中正确的是：C）p=&a2. 设已定义“int x，*p=&x；”，则下列表达式中错误的是：B）&*x3. 若已定义“int a=1 ,*b=&a;”，则“printf（“%d \n”,*b）;”的输出结果为：A）a的值。

4. 设已定义“int x，*p，*pl=&x，*p2=&x;”,则下列表达式中错误的是：C）p=p1+p2.5. 设有函数定义“void p(int *x){printf(“%d\n”,*x);}”和变量定义“int a=3;”,则正确的函数调用是：C）p（&a）6. 函数“int fun（char * x）{char * y=x; while(*y)y++;return(y-x); }”的功能是A）求字符串的长度。

7. 运行一下程序，输出结果为：B）5 6int fun (int a,int *b){a++；（*b）++；return a+*b；}void main(){int x=1,y=2;Printf(“%d”,fun(x,&y));Printf(“%d”,fun(x,&y));}8. 运行以下程序，输出结果为：C）58#include<stdio.h>Int * fun(int a ,int *b){a++;(*b)++;*b=a+*b;return b;}Void main(){Int x=1,y=2,*z;Z=fun(x,&y);Printf(“%d”,*z);Z=fun(x,&y);Printf(“%d”,*z);}9. 若已定义“int a[]={1,2 ,3,4},*p=a;”,则下面表达式中值不等于2的是C）*(++a)10. 若已定义“int a[]={1,2 ,3,4},*p=a+1;”,则p[2]的值为C）411. 设已定义“int x[4][10],*p=x[0];”,则下列表达式中的值为整形的是B）*（p+1）12. 设已定义“char s[]=”ABCD”;”,”printf(“%s”,s+1)”的值为C）BCD13. 设已定义“char str[]=”abcd”,*ptr=str;”,则*（prt+4）的值为B）014. 下面对字符串变量的初始化或赋值操作中，错误的是C）char a[10];a=”OK”;15. 设已定义“char *ps[2]={“abc”,”1234”};”,则以下叙述中错误的是A）ps为指针变量，它指向一个长度为2的字符串数组16. 设已定义“struct {int a,b;} s,*ps=&s;”,则错误的结构体成员引用是C）*ps.a17. 设已有以下定义，则表达式的值为2的是A）k=++p->datastruct st {int data;st *link;} a[3]={1,a+1,3,a+2,5,0},*p=a;二、编程题1. 输入3个字符串，输出其中最大的字符串（用字符指针）#include <stdio.h>#include <stdlib.h>#define str_count 3#define str_length 100int main(int argc, char *argv[]){char a[str_count][str_length],*p;printf("请输入3个字符串：");int i;for(i=0;i<str_count;i++){scanf("%s",a[i]);}p=a[0];for(i=1;i<str_count;i++){if(strcmp(p,a[i])<0){p=a[i];}}printf("最大的字符串为:%s",p);system("PAUSE");return 0;}2. 定义一个函数，函数的功能是求已知半径的圆的周长和面积。

c程序设计第四版(谭浩强)第十章答案

c程序设计第四版(谭浩强)第十章答案
}employee;
int main()
{
employee em[10];
FILE *p;
int i;
if((p=fopen("employee.dat","wb"))==NULL)
stu[n-1].ave=(stu[n-1].score[0]+stu[n-1].score[1]+stu[n-1].score[2])/3.0;
sort(stu);
if((p1=fopen("stud2.dat","wb"))==NULL)
{
printf("cannot open file stud\n");
}
for(i=0;i<n;i++)
if(fwrite(&stu[i],sizeof(struct student),1,p)!=1)
printf("file write error\n");
fclose(p);
}
int main()
{
int i;
printf("enter date of students:\n");
exit(0);
}
printf("new sort of students:\n");
for(i=0;i<n;i++)
{
fwrite(&stu[i],sizeof(struct student),1,p1);
printf("%ld %-6s %5.2f %5.2f %5.2f %5.2f\n",stu[i].num,stu[i].name,stu[i].score[0],stu[i].score[1],stu[i].score[2],stu[i].ave);

谭浩强版C语言的第十章《指针》答案

谭浩强版C语言的第十章《指针》答案

谭浩强版C语言的第十章《指针》答案第十章《指针》答案如下inc/testPtr.h#include <string.h>#include <ctype.h>#include <math.h>#include <assert.h>#define SIZE 1024int a2i(char *start, char *end){int size = 0, ret = 0;long base = 0;size = end - start + 1;base = (long)pow(10, size - 1);while(size-- > 0){ret += (*start++ - '0') * base;base /= 10;}return ret;}int extraNum(char *str, int arr[]){int ite = 0, counter = 0;char *start = NULL, *end = NULL;while(*str != '\0'){if(isdigit(*str)){start = end = str;while( isdigit(*end) && *end != '\0'){++end;}arr[ite++] = a2i(start, end-1);str = end;}else{str++;}}return ite;}int sortStr(char *arr[], int size){int ite1 = 0, ite2 = 0, minPos = 0;char *tmp;for(ite1 = 0; ite1 < size - 1; ite1++){minPos = ite1;for(ite2 = ite1 + 1; ite2 < size; ++ite2 ){if( strcmp(arr[ite2], arr[minPos]) < 0 ){minPos = ite2;}}if(minPos != ite1){tmp = arr[minPos];arr[minPos] = arr[ite1];arr[ite1] = tmp;}}return 0;}int sort(int arr[], int size){int minPos = 0, ite1 = 0, ite2 = 0, tmp = 0;for(ite1 = 0; ite1 < size - 1; ite1++){minPos = ite1;for(ite2 = ite1 + 1; ite2 < size; ite2++){if( arr[ite2] < arr[minPos] ){minPos = ite2;}}if(minPos != ite1){tmp = arr[ite1];arr[ite1] = arr[minPos];arr[minPos] = tmp;}}return 0;}int sortPtr(int arr[], int size){int minPos = 0, ite1 = 0, ite2 = 0, tmp = 0;for(ite1 = 0; ite1 < size - 1; ite1++){minPos = ite1;for(ite2 = ite1 + 1; ite2 < size; ite2++){if( *(arr + ite2) < *(arr + minPos) ){minPos = ite2;}}if(minPos != ite1){tmp = *(arr + ite1);*(arr + ite1) = *(arr + minPos);*(arr + minPos) = tmp;}}return 0;}int getMultiArr(int arr[][5], int n){int i = 0, j = 0, min = 0, tmp = 0, size = 5 * n;int *p = NULL, pos[size];/* copy */p = (int *)malloc(size * sizeof(int));assert(p != NULL);memcpy(p, arr, size * sizeof(int));/* sort */for(i = 0; i < size - 1; i++){min = i;for(j = i + 1; j < size; j++){if( *(p + j) < *(p + min)){min = j;}}if(min != i){tmp = *(p + min);*(p + min) = *(p + i);*(p + i) = tmp;}}/* move */for(i = 0; i < n; i++){for(j = 0; j < 5; j++){if( *p == arr[i][j] ){tmp = arr[i][j];arr[i][j] = arr[0][0];arr[0][0] = tmp;continue;}if( *(p + 1) == arr[i][j] ){tmp = arr[i][j];arr[i][j] = arr[0][4];arr[0][4] = tmp;continue;}if( *(p + 2) == arr[i][j] ){tmp = arr[i][j];arr[i][j] = arr[n - 1][0];arr[n - 1][0] = tmp;continue;}if( *(p + 3) == arr[i][j] ){tmp = arr[i][j];arr[i][j] = arr[n - 1][4];arr[n - 1][4] = tmp;continue;}if( *(p + size - 1) == arr[i][j] ){tmp = arr[i][j];arr[i][j] = arr[n/2][2];arr[n/2][2] = tmp;continue;}}}free(p);p = NULL;return 0;}int statStr(char *str){int upper = 0, lower = 0, space = 0, num = 0, other = 0;while(*str != '\0'){if(isdigit(*str)){num++;}else if (isupper(*str)){upper++;}else if (islower(*str)){lower++;}else if (isspace(*str)){space++;}else{other++;}str++;}assert(3 == upper);assert(5 == lower);assert(10 == num);assert(2 == space);assert(6 == other);return 0;}int average(int(*stu)[6], int classNum, int stuNum){int i = 0, ave = 0;for(i = 0; i < stuNum; i++){ave += (*(stu + i))[classNum];}ave /= stuNum;return ave;}int searchStu(int(*stu)[6], int stuNum){int counter = 0, i = 0, j = 1, stuCounter = 0;for (i = 0; i < stuNum; i++){counter =0;for(j = 1; j < 6; j++){if( (*(stu + i))[j] < 60 ){counter++;}}if(counter >= 2){stuCounter++;}}return stuCounter;}int moveInt(int arr[], int n, int m){int i = 0, *p = NULL;p = (int*)malloc(n * sizeof(int));assert(p != NULL);memcpy(p + m, arr, (n - m) * sizeof(int));memcpy(p, arr + n -m , m * sizeof(int));memcpy(arr, p, n * sizeof(int));free(p);p = NULL;return 0;}int myStrcmp(char *p1, char*p2){int ret = 0;while((*p1 != '\0') && (*p2 != '\0') && ( *p1 == *p2 ) ) {p1++;p2++;}if(*p1 == '\0'){ret = -1;}else if (*p2 == '\0'){ret = 1;}else{ret = (*p1 - *p2) > 0 ? 1 : -1;}return ret;}int revArr(int a[], int size){int tmp = 0, *start = NULL, *end = NULL;start = a;end = start + size - 1;size = size / 2;while(size >= 0){tmp = *(start + size);*(start + size) = *(end - size);*(end - size) = tmp;size--;}return 0;}char *getMonth(char *month[], int which) {assert(which <= 12);return ( *(month + which - 1));}int getStr(char *dest, char* src, int m) {int len = 0;len = strlen(src) + 1 - m;src = src + m - 1;memcpy(dest, src, len * sizeof(char));return 0;}int removePer3(int arr[], int size){int i = 0;for(i = 0; i < size; i++){if(((arr[i]) % 3) == 0){arr[i] = 0;}}return 0;}int getMinMax(int a[], int size){int i = 0, min = 0, max = 0, tmp = 0;min = max = 0;for(i = 0; i < size; i++){if (a[i] <= a[min]){min = i;}if(a[i] >= a[max]){max = i;}}tmp = a[0];a[0] = a[min];a[min] = tmp;tmp = a[size - 1];a[size - 1] = a[max];a[max] = tmp;}int test_10_1(){int ite = 0, iRet = 0, arr[5] = {121, 234, 456456, 543, 23};iRet = sortPtr(arr, 5);assert (23 == arr[0]);assert (121 == arr[1]);assert (234 == arr[2]);assert (543 == arr[3]);assert (456456 == arr[4]);printf("\r\nTest_10_1 Passed!");return 0;}int test_10_2(){int ite = 0, iRet = 0;char *arr[10] = { "In the IBM Rational ClearCase environment", \"An auditable history of source files and software builds is maintained in your organization", \"The efforts of your team can be coordinated into a definable"};iRet = sortStr(arr, 3);assert( strcmp(arr[0], "An auditable history of source files and software builds is maintained in your organization") == 0 );assert( strcmp(arr[1], "In the IBM Rational ClearCase environment") == 0);assert( strcmp(arr[2], "The efforts of your team can be coordinated into a definable") == 0);printf("\r\nTest_10_2 Passed!");return 0;}int test_10_3(){int ret = 0, a[10] = {7,2,3,9,1,0,7,6,5,0};ret = getMinMax(a, 10);assert(a[0] == 0);assert(a[9] == 9);printf("\r\nTest_10_3 Passed!");return 0;}int test_10_4(){int ret = 0, arr[9] = {0, 1, 2, 3, 4, 5, 6, 7, 8};ret = moveInt(arr, 9, 3);assert(6 == arr[0]);assert(7 == arr[1]);assert(8 == arr[2]);assert(0 == arr[3]);assert(1 == arr[4]);assert(2 == arr[5]);assert(3 == arr[6]);assert(4 == arr[7]);assert(5 == arr[8]);printf("\r\nTest_10_4 Passed!");return 0;}int test_10_5(){int ret = 0, i = 0, a[12] = {1,2,3,4,5,6,7,8,9,10,11,12};ret = removePer3(a, 12);assert(a[2] == 0);assert(a[5] == 0);assert(a[8] == 0);assert(a[11] == 0);printf("\r\nTest_10_5 Passed!");return 0;}int test_10_7(){int ret = 0;char s2[100] = {0}, *s1 = "hello world!";getStr(s2, s1, 7);assert( strcmp(s2, "world!") == 0);printf("\r\nTest_10_7 Passed!");return 0;}int test_10_8(){int ret = 0;char *str = "a123xABC ??#$%^ 302tab5876";ret = statStr(str);if(ret == 0){printf("\r\nTest_10_8 Passed!");}return 0;}int test_10_10(){int i = 0, j = 0, ret = 0;int arr[5][5] = { \{16, 17, 18, 19, 20}, \{11, 12, 13, 14, 15}, \{1, 2, 3, 4, 5}, \{21, 22, 23, 24, 25}, \{6, 7, 8, 9, 10}, \};ret = getMultiArr(arr, 5);assert(1 == arr[0][0] );assert(2 == arr[0][4] );assert(3 == arr[4][0] );assert(4 == arr[4][4] );assert(25 == arr[2][2] );printf("\r\nTest_10_10 Passed!");return 0;}int test_10_11(){int ite = 0, iRet = 0;char *arr[10] = { "In the", \"An aud", \"The ef", \"Proces", \"Sets o", \"Unifie", \"Out-of", \"Practi", \"Ration", \"Explor" \};iRet = sortStr(arr, 10);assert( strcmp(arr[0], "An aud") == 0 );assert( strcmp(arr[1], "Explor") == 0);assert( strcmp(arr[2], "In the") == 0);assert( strcmp(arr[3], "Out-of") == 0);assert( strcmp(arr[4], "Practi") == 0);assert( strcmp(arr[5], "Proces") == 0);assert( strcmp(arr[6], "Ration") == 0);assert( strcmp(arr[7], "Sets o") == 0);assert( strcmp(arr[8], "The ef") == 0);assert( strcmp(arr[9], "Unifie") == 0);printf("\r\nTest_10_11 Passed!");return 0;}int test_10_12(){int ite = 0, iRet = 0;char *arr[10] = { "In the IBM Rational ClearCase environment", \"An auditable history of source files and software builds is maintained in your organization", \"The efforts of your team can be coordinated into a definable", \"Process by using one of the following", \"Sets of Rational ClearCase features", \"Unified Change Management (UCM),", \"Out-of-the-box process that supports best", \"Practices for change management as described in the IBM", \"Rational Unified Process. Project managers can configure", \"Explorer. For more information about Rational ClearCase"};iRet = sortStr(arr, 10);assert( strcmp(arr[0], "An auditable history of source files and software builds is maintained in your organization") == 0 );assert( strcmp(arr[1], "Explorer. For more information about Rational ClearCase") == 0);assert( strcmp(arr[2], "In the IBM Rational ClearCase environment") == 0);assert( strcmp(arr[3], "Out-of-the-box process that supports best") == 0);assert( strcmp(arr[4], "Practices for change management as described in the IBM") == 0);assert( strcmp(arr[5], "Process by using one of the following") == 0);assert( strcmp(arr[6], "Rational Unified Process. Project managers can configure") == 0);assert( strcmp(arr[7], "Sets of Rational ClearCase features") == 0);assert( strcmp(arr[8], "The efforts of your team can be coordinated into a definable") == 0);assert( strcmp(arr[9], "Unified Change Management (UCM),") == 0);printf("\r\nTest_10_12 Passed!");return 0;}int test_10_14(){int ret = 0, a[11] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};ret = revArr(a, 11);assert(10 == a[0]);assert(9 == a[1]);assert(8 == a[2]);assert(7 == a[3]);assert(6 == a[4]);assert(5 == a[5]);assert(4 == a[6]);assert(3 == a[7]);assert(2 == a[8]);assert(1 == a[9]);assert(0 == a[10]);printf("\r\nTest_10_14 Passed!");return 0;}int test_10_15(){int ave = 0, ret = 0;int student[4][6] = { \{1, 100, 90, 80, 70, 97}, \{2, 34, 45, 56, 78, 97}, \{3, 76, 34, 68, 84, 12}, \{4, 90, 90, 90, 75, 28} \};ave = average(student, 1, 4);ret = searchStu(student, 4);assert(75 == ave);assert(2 == ret);printf("\r\nTest_10_15 Passed!");return 0;}int test_10_16(){int iRet = 0, ite = 0, arr[SIZE] = {0};char *str = "a123x456 17960? 302tab5876";iRet = extraNum(str, arr);assert(123 == arr[0]);assert(456 == arr[1]);assert(17960 == arr[2]);assert(302 == arr[3]);assert(5876 == arr[4]);printf("\r\nTest_10_16 Passed!");return 0;}int test_10_17(){int ret = 0;char *s1 = "abcd", *s2 = "abCd";ret = myStrcmp(s1, s2);assert(ret == 1);char *s3 = "aBcd", *s4 = "abCd";ret = myStrcmp(s3, s4);assert(ret == -1);char *s5 = "abcde", *s6 = "abcd";ret = myStrcmp(s5, s6);assert(ret == 1);char *s7 = "abcd", *s8 = "abcde";ret = myStrcmp(s7, s8);assert(ret == -1);char *s9 = "abcd", *s10 = "abCde";ret = myStrcmp(s9, s10);assert(ret == 1);printf("\r\nTest_10_17 Passed!");return 0;}int test_10_18(){int which = 0;char *month[12] = { \"January", \"February", \"March", \"April", \"May", \"June", \"July", \"August", \"September", \"October", \"November", \"December" \};which = 11;assert( strcmp("November", getMonth(month, which)) == 0 );which = 7;assert( strcmp("July", getMonth(month, which)) == 0 );printf("\r\nTest_10_18 Passed!");return 0;}int test_10_20(){int ite = 0, iRet = 0;char *arr[5] = { "In the IBM Rational ClearCase environment", \"An auditable history of source files and software builds is maintained in your organization", \"The efforts of your team can be coordinated into a definable", \"Process by using one of the following", \"Sets of Rational ClearCase features" \};iRet = sortStr(arr, 5);assert( strcmp(arr[0], "An auditable history of source files and software builds is maintained in your organization") == 0 );assert( strcmp(arr[1], "In the IBM Rational ClearCase environment") == 0);assert( strcmp(arr[2], "Process by using one of the following") == 0);assert( strcmp(arr[3], "Sets of Rational ClearCase features") == 0);assert( strcmp(arr[4], "The efforts of your team can be coordinated into a definable") == 0);printf("\r\nTest_10_20 Passed!");return 0;}int test_10_21(){int ite = 0, iRet = 0, arr[5] = {121, 234, 456456, 543, 23};iRet = sort(arr, 5);assert (23 == arr[0]);assert (121 == arr[1]);assert (234 == arr[2]);assert (543 == arr[3]);assert (456456 == arr[4]);printf("\r\nTest_10_21 Passed!");return 0;}int testPtr(){int iRet = 0;#if 0#endifiRet += test_10_1();iRet += test_10_2();iRet += test_10_3();iRet += test_10_4();iRet += test_10_5();iRet += test_10_7();iRet += test_10_8();iRet += test_10_10();iRet += test_10_11();iRet += test_10_12();iRet += test_10_14();iRet += test_10_15();iRet += test_10_16();iRet += test_10_17();iRet += test_10_18();iRet += test_10_20();iRet += test_10_21();return iRet;}src/#include <stdio.h>#include <stdlib.h>#include <assert.h>#include "../inc/testFile.h" #include "../inc/testBits.h" #include "../inc/testPtr.h"int main(){int iRet = 0;#if 0iRet += testFile();assert(iRet == 0);iRet += testBits();assert(iRet == 0);#endifiRet += testPtr();assert(iRet == 0);return 0;}。

C语言习题结构体和杂类(答案)

C语言习题结构体和杂类(答案)

C语言习题结构体和杂类（答案）第十章结构体和杂类一.选择题1.如下说明语句，则下面叙述不正确的是（C）。

tructtu{inta;floatb;}tutype;A.truct是结构体类型的关键字B.tructtu是用户定义结构体类型C.tutype是用户定义的结构体类型名(变量名)D.a和b都是结构体成员名2.在16位PC机中，若有定义：tructdata{inti;charch;doublef;}b;则结构变量b占用内存的字节数是（D）。

A.1B.2C.8D.11A.1和2B.2和3C.7和2D.7和84.以下程序的输出结果是(D)。

unionmyun{truct{int某,y,z;}u;intk;}a;main(){a.u.某=4;a.u.y=5;a.u.z=6;a.k=0;printf(\A.4B.5C.6D.05.当定义一个共用体变量时，系统分配给它的内存是(C)。

A.各成员所需内存量的总和B.结构中第一个成员所需内存量C.成员中占内存量最大的容量D.结构中最后一个成员所需内存量6.若有以下程序段：uniondata{inti;charc;floatf;}a;intn;则以下语句正确的是(C)。

A.a=5;B.a={2,’a’,1.2}C.printf(“%d”,a);D.n=a;7.设truct{inta;charb;}Q,某p=&Q;错误的表达式是（d）。

A.Q.aB.(某p).bC.p->aD.某p.b9.以下对C语言中共用体类型数据的叙述正确的是（c）。

A.可以对共用体变量直接赋值B.一个共用体变量中可以同时存放其所有成员C.一个共用体变量中不能同时存放其所有成员D.共用体类型定义中不能出现结构体类型的成员10.下面对typedef的叙述中不正确的是（b）。

A.用typedef可以定义多种类型名，但不能用来定义变量B.用typedef可以增加新类型C.用typedef只是将已存在的类型用一个新的标识符来代表D.使用typedef有利于程序的通用和移植二.判断题1.共用体类型的变量的字节数等于各成员字节数之和。

C语言-谭浩强著-第10章练习题

C语言-谭浩强著-第10章练习题

第10章对文件的输入输出（2012年9月真题）（40）有以下程序#include <stdio.h>main(){ FILE *fp;int i,a[6]={1,2,3,4,5,6};fp=fopen("d2.dat","w+");for(i=0;i<6;i++) fprintf(fp,"%d\n",a[i]);rewind(fp);for(i=0;i<6;i++) fscanf(fp,"%d",&a[5-i]);fclose(fp);for(i=0;i<6;i++) printf("%d,",a[i]);}程序运行后的输出结果是A）4,5,6,1,2,3, B）1,2,3,3,2,1,C）1,2,3,4,5,6, D）6,5,4,3,2,1,答案：D（2012年3月真题）40、以下函数不能用于向文件写入数据的是A ftellB fwriteC fputcD fprintf答案：A（2011年9月真题）40.有以下程序#include<stdio.h>main(){FILE *fp; int k,n,i,a[6]={1,2,3,4,5,6};fp=fopen("d2.dat","w");for(i=0;i<6;i++)fprintf(fp,"%d\n",a[i]);fclose(fp); fp=fopen("d2.dat","r");for(i=0;i<3;i++)fscanf(fp,"%d%d",&k,&n);fclose(fp); printf("%d,%d\n",k,n); }程序运行后的输出结果是A.1,2B.3,4C.5,6D.123,456答案：C（2011年3月真题）(40)设fp已定义，执行语句fp=fopen("file","w");后，以下针对文本文件file操作叙述的选项中正确的是A)写操作结束后可以从头开始读 B)只能写不能读C)可以在原有内容后追加写 D)可以随意读和写（2010年3月真题）(40)有以下程序#include#include<stdio.h>main(){ FILE *fp;char str[10];fp=fopen("myfile.dat","w");fputs("abc",fp);fclose(fp);fopen("myfile.data","a+");fprintf(fp,"%d",28);rewind(fp);fscanf(fp,"%s",str); puts(str);fclose(fp); }程序运行后的输出结果是A)abc B) 28c C) abc28 D)因类型不一致而出错答案：B（2009年9月真题）（40）下列关于C语言文件的叙述中正确的是A）文件由一系列数据依次排列组成，只能构成二进制文件B）文件由结构序列组成，可以构成二进制文件或文本文件C）文件由数据序列组成，可以构成二进制文件或文本文件D）文件由字符序列组成，其类型只能是文本文件答案：C（2009年3月真题）40.有以下程序#include <stdio.h>main(){ FILE *f;f=fopen("filea.txt","w"); fprintf(f,"abc"); fclose(f); }若文本文件filea.txt中原有内容为:hello,则运行以上程序后,文件filea.txt的内容为A)helloabc B)abclo C)abc D)abchello答案：C（2008年9月真题）（40）有以下程序#include <stdio.h>main(){ FILE *pf;char *s1="China", *s2="Beijing";pf=fopen("abc.dat","wb+");fwrite(s2,7,1,pf);rewind(pf); /*文件位置指针回到文件开头*/fwrite(s1,5,1,pf);fclose(pf);}以上程序执行后abc.dat文件的内容是A) China B) Chinang C) ChinaBeijing D) Bei jingChina答案：B（2008年4月真题）30、下列叙述中错误的是( )。

C语言习题及答案(第十章)

C语言习题及答案(第十章)

10-2 定义指针变量p,q,r,让它们指向变量a,b,c,在指向d,e,f,最后指向变量x,y,z,然后输出p,q,r与*p,*q,*r.解：int a,b,c,d,e,f,x,y,z ;int *p,*q,*r ;p=&a;q=&b;r=&c ;p=&d;q=&e;r=&f ;p=&x;q=&y;r=&z ;printf(“%l,%l,%l”,p,q,r) ;printf(“%d,%d,%d”,*p,*q,*r) ;思考：指针也是一种数据类型吗？他可以不依赖其它数据类型而独立存在吗？10-3 应用指针，实现10个整数从打到小的排序输出。

解：#include "stdio.h"void main(){int i,j,a[10],*p,t;printf("请输入10个整数：\n");for(i=0;i<10;i++)scanf("%d",&a[i]);for(i=0;i<9;i++){p=&a[i] ;for(j=i+1;j<10;j++)if(a[j]>*p) p=&a[j] ;t=*p;*p=a[i];a[i]=t;}printf("排序后的数组为：\n");for(i=0;i<10;i++)printf("%d,",a[i]) ;}思考：指针变量的值可以由键盘输入确定吗？它与其所指类型的变量值之间的关系如何？10-4 应用指针，求n个数的最小值和最大值。

解：#include "stdio.h"#define L 100void main(){int i,n,a[L],*max,*min;printf("请确定欲输入数据的个数（<100）：");scanf("%d",&n);printf("下面请依次输入%d的个数\n",n);for(i=0;i<n;i++){printf("第%d的个数：",i+1);scanf("%d",&a[i]);}max=min=&a[0] ;for(i=0;i<n;i++){if(a[i]>*max) max=&a[i];if(a[i]<*min) min=&a[i];}printf("最大值=%4d,最小值=%4d\n",*max,*min);}思考：指针也可以指向相同类型的数组元素，此题如用指针来表示来表示数组的各元素的值，程序应如何设计？10-5 应用指针，编写下列字符串处理函数：(1) 字符串的复制函数。

明解C语言入门篇第十章答案

明解C语言入门篇第十章答案

明解C语⾔⼊门篇第⼗章答案练习10-1#include <stdio.h>void adjust_point(int*n) {if (*n > 100)*n = 100;if (*n < 0)*n = 0;}int main() {int x;printf("请输⼊⼀个数：");scanf("%d", &x);adjust_point(&x);printf("修改后的值是%d",x);}练习10-2void decrement_date(int* y, int* m, int* d) {*d -= 1;if (*d == 0) {*m -= 1;if (*m == 1 || *m == 3 || *m == 5 || *m == 7 || *m == 10 || *m == 8 ||* m == 0) {*d = 31;}if (*m == 2) {*d = 28;if (*y % 4 == 0) {*d = 29;}}else*d = 30;if (*m == 0) {*m = 12;*y -= 1;}}}void increment_date(int* y, int* m, int* d) {*d += 1;if ((*m == 1 || *m == 3 || *m == 5 || *m == 7 || *m == 10 || *m == 8 || *m == 12) && (*d == 32)) {*d = 1;*m += 1;}if ((*m == 4 || *m == 6 || *m == 9 || *m == 11 ) && (*d == 31)) {*d = 1;*m += 1;}if (*m == 2) {if (*d == 29&& ((*y & 4) != 0)) {*d = 1;}if (*d == 30 && ((*y & 4) == 0)) {*d = 1;*m += 1;}}if (*m == 13) {*y += 1;*m = 1;}}练习10-3#include<stdio.h>void swap(int* px, int*py) {int temp = *px;*px = *py;*py = temp;}void sort3(int*n1, int*n2, int*n3) {if (*n1 > *n2) {swap(n1, n2);}if (*n1 > * n3) {swap(n1, n3);}if (*n2 > * n3) {swap(n2, n3);}}int main(void) {int n1, n2, n3;printf("n1=");scanf("%d", &n1);printf("n2=");scanf("%d", &n2);printf("n3=");scanf("%d", &n3);putchar('\n');sort3(&n1, &n2, &n3);printf("%d,%d,%d", n1, n2, n3); }练习10-4#include <stdio.h>#define number 5void set_idx(int*v, int n) {int i = 0;for (i = 0; i < n; i++) {*(v + i) = i;}}int main() {int i;set_idx(x, number);for (i = 0; i < number; i++) {printf("x[%d]=%d", i,x[i] );putchar('\n');}}练习10-5会报错数组中元素会溢出，因为直接从v【2】开始赋值。

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C程序设计[谭浩强]（第三版）课后答案第十章指针2010-06-25 14:04第十章指针10.1main(){int n1,n2,n3;int *p1,*p2,*p3;scanf("%d,%d,%d",&n1,&n2,&n3);p1=&n1;p2=&n2;p3=&n3;if(n1>n2)swap(p1,p2);if(n1>n3)swap(p1,p3);if(n2>n3)swap(p2,p3);printf("%d,%d,%d\n",n1,n2,n3);}swap(p1,p2)int *p1,*p2;{int p;p=*p1;*p1=*p2;*p2=p;}10.2main(){char *str1[20],*str2[20],*str3[20];char swap();scanf("%s",str1);scanf("%s",str2);scanf("%s",str3);if(strcmp(str1,str2)>0)swap(str1,str2);if(strcmp(str1,str3)>0)swap(str1,str3);if(strcmp(str2,str3)>0)swap(str2,str3);printf("%s\n%s\n%s\n",str1,str2,str3);}char swap(p1,p2)char *p1,*p2;{char *p[20];strcpy(p,p1);strcpy(p1,p2);strcpy(p2,p);}10.3main(){int number[10];input(number);max_min_value(number); output(number);}input(number)int number[10];{int i;for(i=0;i<10;i++)scanf("%d",&number);}max_min_value(number)int number[10];{int *max,*min;int *p,*end;end=number+10;max=min=number;for(p=number+1;p<end;p++) if(*p>*max)max=p;else if(*p<*min)min=p;*p=number[0];number[0]=*min;*min=*p;*p=number[9];number[9]=*max;*max=*p;return;}output(number)int number[10];{int *p;for(p=number;p<number+9;p++) printf("%d,",*p);printf("%d\n",*p);}10.4main(){int number[20],n,m,i;scanf("%d",&n);scanf("%d",&m);for(i=0;i<n:i++)scanf("%d",&number); move(number,n,m);for(i=0;i<n;i++)printf("%8d",number);}move(array,n,m)int array[20],n,m;{int *p,end;end=*(array+n-1);for(p=array+n-1;p>array;p--) *p=*(p-1);*array=end;m--;if(m>0)move(array,n,m);}10.5#define nmax 50main(){int i,k,m,n,num[nmax],*p; scanf("%d",&n);p=num;for(i=0;i<n;i++)*(p+i)=i+1;i=k=m=0;while(m<n-1){if(*(p+i)!=0)k++;if(k==3){*(p+i)=0;k=0;m++;}i++;if(i==n)i=0;}while(*p==0)p++;printf("%d",*p);}10.6main(){int len;char *str[20];scanf("%s",str);len=length(str);printf("\nlen=%d\n",len);}length(p)char *p;{int n=0;while(*p!='\0')return(n);}10.7main(){int m;char *str1[20],*str2[20];scanf("%s",str1);scanf("%d",&m);if(strlen(str1)<m)printf("error");else{copystr(str1,str2,m);printf("%s",str2);}}copystr(p1,p2,m)char *p1,*p2;int m;{int n=0;while(n<m-1){n++;p1++;}while(*p1!='\0'){*p2=*p1;p1++;p2++;}*p2='\0';}10.8#include"stdio.h"main(){int cle=0,sle=0,di=0,wsp=0,ot=0,i; char *p,s[20];for(i=0;i<20;i++)s=0;i=0;while((s=getchar())!='\n')i++;p=s;while(*p!='\n'){if(*p>='a'&&*p<='z')++sle;else if(*p>='A'&&*p<='Z')++cle;else if(*p==' ')else if(*p>='0'&&*p<='9')++di;else++ot;p++;}printf("sle=%d,cle=%d,wsp=%d,di=%d,ot=%d\n",sle,cle,wsp,di,ot); }10.9main(){int a[3][3],*p,i;for(i=0;i<3;i++)scanf("%d,%d,%d",a[0],a[1],a[2]);p=a;move(p);for(i=0;i<3;i++)printf("%d %d %d\n",a[0],a[1],a[2]);}move(pointer)int *pointer;{int i,j,t;for(i=0;i<2;i++)for(j=i+1;j<3;j++){t=*(pointer+3*i+j);*(pointer+3*i+j)=*(pointer+3*j+i);*(pointer+3*j+i)=t;}}10.10main(){int a[5][5],*p,i,j;for(i=0;i<5;i++)for(j=0;j<5;j++)scanf("%d",&a[j]);p=a;change(p);for(i=0;i<5;i++){printf("\n");for(j=0;j<5;j++)printf("%8d",a[j]);}}change(p)int *p;{int i,j,change;int *pmax,*pmin;pmax=p;pmin=p;for(i=0;i<5;i++)for(j=0;j<5;j++){if(*pmax<*(p+5*i+j))pmax=p+5*i+j;if(*pmin>*(p+5*i+j))pmin=p+5*i+j;}change=*(p+12);*(p+12)=*pmax;*pmax=change;change=*p;*p=*pmin;*pmin=change;pmin=p+1;for(i=0;i<5;i++)for(j=0;j<5;j++)if(((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j; change=*(p+4);*(p+4)=*pmin;*pmin=change;pmin=p+1;for(i=0;i<5;i++)for(j=0;j<5;j++)if(((p+5*i+j)!=(p+4))&&((p+5*i+j)!=p)&&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;change=*(p+20);*(p+20)=*pmin;*pmin=change;pmin=p+1;for(i=0;i<5;i++)for(j=0;j<5;j++)if(((p+5*i+j)!=p)&&((p+5*i+j)!=(p+4))&&((p+5*i+j)!=(p+20)) &&(*pmin>*(p+5*i+j)))pmin=p+5*i+j;change=*(p+24);*(p+24)=*pmin;*pmin=change;}10.11main(){int i;char *p,str[10][10];for(i=0;i<10;i++)scanf("%s",str);p=str;sort(p);for(i=0;i<10;i++)printf("%s\n",str);}sort(p)char *p;{int i,j;char s[10],*smax,*smin;for(i=0;i<10;i++){smax=p+10*i;for(j=i+1;j<10;j++){smin=p+10*j;if(strcmp(smax,smin)>0){strcpy(s,smin);strcpy(smin,smax);strcpy(smax,s);}}}}10.12#define MAX 20main(){int i;char *pstr[10],str[10][MAX];for(i=0;i<10;i++)pstr=str;for(i=0;i<10;i++)scanf("%s",pstr);sort(pstr);for(i=0;i<10;i++)printf("%s\n",pstr);}sort(pstr)char *pstr[10];{int i,j;char *p;for(i=0;i<10;i++){for(j=i+1;j<10;j++){if(strcmp(*(pstr+i),*(pstr+j))>0){p=*(pstr+i);*(pstr+i)=*(pstr+j);*(pstr+j)=p;}}}}10.13#include"math.h"main(){int n=20;float a,b,a1,b1,a2,b2,c,(*p)(),jiff(); scanf("%f,%f",&a,&b);scanf("%f,%f",&a1,&b1);scanf("%f,%f",&a2,&b2);p=sin;c=jiff(a,b,n,p);printf("sin=%f\n",c);p=cos;c=jiff(a1,b1,n,p);printf("cos=%f\n",c);p=exp;c=jiff(a2,b2,n,p);printf("exp=%f\n",c);}float jiff(a,b,n,p)float a,b,(*p)();int n;{int i;float x,f,h,area;h=(b-a)/n;x=a;area=0;for(i=1;i<=n;i++){x=x+h;area=area+(*p)(x)*h;}return(area);}10.14main(){int i,n,num[20];char *p;scanf("%d",&n);for(i=0;i<n;i++)scanf("%d",&num);p=num;sort(p,n);for(i=0;i<n;i++)printf("%8d",num);}sort(p,m)char *p;int m;{int i;char change,*p1,*p2;for(i=0;i<m/2;i++){p1=p+i;p2=p+(m-1-i);change=*p1;*p1=*p2;*p2=change;}}10.15main(){int i,j,*pnum,num[4];float score[4][5],aver[4],*psco,*pave; char course[5][10],*pcou;pcou=course[0];for(i=0;i<5;i++)scanf("%s",pcou+10*i);printf("number");for(i=0;i<5;i++)printf(",%s",pcou+10*i);printf("\n");psco=score;pnum=num;for(i=0;i<4;i++){scanf("%d",pnum+i);for(j=0;j<5;j++)scanf(",%f",psco+5*i+j);}pave=aver;printf("\n");avsco(psco,pave);avcour1(pcou,psco);printf("\n");fali2(pcou,pnum,psco,pave);printf("\n");good(pcou,pnum,psco,pave);}avsco(psco,pave)float *psco,*pave;{int i,j;float sum,average;for(i=0;i<4;i++){sum=0;for(j=0;j<5;j+)sum+=(*(psco+5*i+j));average=sum/5;*(pave+i)=average;}}avcour1(pcou,psco)char *pcou;float *psco;{int i;float sum,average1;sum=0;for(i=0;i<4;i++)sum+=(*(psco+5*i))average1=sum/4;printf("%s %5.2f\n",pcou,average1); }fali2(pcou,pnum,psco,pave)char *pcou;int *pnum;float *psco,*pave;{int i,j,k,label;printf("\nnumber\n");for(i=0;i<5;i++)printf("%-8s",pcou+10*i);printf("\naverage\n");for(i=0;i<4;i++){label=0;for(j=0;j<5;j++)if(*(psco+5*i+j)<60.0)label++;if(label>=2){printf("%-8d",*(pnum+i));for(k=0;k<5;k++)printf("%-8.2f",*(psco+5*i+k));printf("%-8.2f",*(pave+i));}}}good(pcou,pnum,psco,pave)char *pcou;int *pnum;float *psco,*pave;{int i,j,k,label;printf("number");for(i=0;i<5;i++)printf("%-8s",pcou+10*i);printf("average");for(i=0;i<4;i++){label=0;for(j=0;j<5;j++)if(*(psco+5*i+j)>=85.0)label++;if((label>=5)||(*(pave+i)>=90)){printf("%-8d",*(pnum+i));for(k=0;k<5;k++)printf("%-8.2f",*(psco+5*i+k));printf("%-8.2f",*(pave+i));}}}10.16#include"stdio.h"main(){char str[50],*pstr;int i,j,k,m,e10,digit,ndigit,a[10],*pa;gets(str);pstr=str;pa=a;ndigit=0;i=j=0;while(*(pstr+i)!='\0'){if((*(pstr+i)>='0')&&(*(pstr+i)<='9')) j++;else{if(j>0){digit=*(pstr+i-1)-48;k=1;while(k<j){e10=1;for(m=1;m<=k;m++)e10=e10*10;digit+=(*(pstr+i-1-k)-48)*e10;k++;}*pa=digit;ndigit++;pa++;j=0;}}i++;}if(j>0){digit=*(pstr+i-1)-48;k=1;while(k<j){e10=1;for(m=1;m<=k;m++)e10=e10*10;digit+=(*(pstr+i-1-k)-48)*e10;k++;}*pa=digit;ndigit++;j=0;}printf("ndigit=%d\n",ndigit);j=0;pa=a;for(j=0;j<ndigit;j++)printf("%d",*(pa+j));}10.17main(){int m;char str1[20],str2[20],*p1,*p2;scanf("%s",str1);scanf("%s",str2);p1=str1;p2=str2;m=strcmp(p1,p2);printf("%d\n",m);}strcmp(p1,p2)char *p1,*p2;{int i=0;while(*(p1+i)==*(p2+i))if(*(p+i++)=='\0')return(0);return(*(p1+i)-*(p2+i));}10.18main(){static char *mname[13]={"illeagl","January","February","March", "April","May","June","July","August","September","October", "November","December"};int n;scanf("%d",&n);if((n>=1)&&(n<=12))printf("%s\n",*(mname+n));elseprintf("error");}10.20main(){int i;char **p,*pstr[5],str[5][10];for(i=0;i<5;i++)pstr=str;for(i=0;i<5;i++)scanf("%s",pstr);p=pstr;sort(p);for(i=0;i<5;i++)printf("%s\n",pstr);}sort(p)char **P;{int i,j;char *pchange;for(i=0;i<5;i++){for(j=i+1;j<5;j++){if(strcmp(*(p+i),*(p+j))>0){pchange=*(p+i);*(p+i)=*(p+j);*(p+j)=pchange;}}}}10.21main(){int i,n,digit[20],**p,*pstr[20];scanf("%d",&n);for(i=0;i<n;i++)pstr=&digit;for(i=0;i<n;i++)scanf("%d",pstr);p=pstr;sort(p,n);for(i=0;i<n;i++)printf("%d ",*pstr);}sort(p,n)int **p,n;{int i,j,*pchange;for(i=0;i<n;i++){for(j=i+1;j<n;j++){if(**(p+i)>**(p+j)){pchange=*(p+i);*(p+i)=*(p+j);*(p+j)=pchange;}}}}第十一章结构体与共用体11.1struct{int year;int month;int day;}date;main(){int days;scanf("%d,%d,%d",&date.year,&date.month,&date.day); switch(date.month){case 1:days=date.day;break;case 2:days=date.day+31;break;case 3:days=date.day+59;break;case 4:days=date.day+90;break;case 5:days=date.day+120;break;case 6:days=date.day+151;break;case 7:days=date.day+181;break;case 8:days=date.day+212;break;case 9:days=date.day+243;break;case 10:days=date.day+273;break;case 11:days=date.day+304;break;case 12:days=date.day+334;break;}if((date.year%4==0&&date.year%100!=0||date.year%400==0)&&date.month>=3)days+=1;printf("days=%d\n",days);}11.2struct dt{int year;int month;int day;}date;main(){scanf("%d,%d,%d",&date.year,&date.month,&date.day); printf("\n%d\n",days(date.year,date.month,date.day));}days(year,month,day)int year,month,day;{int daysum=0,i;static int daytab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31} for(i=1;i<month;i++)daysum+=daytab;daysum+=day;if((year%4==0&&year%100!=0||year%400==0)&&month>=3) daysum+=1;return(daysum);}11.311.4#define N 5struct student{char num[6];char name[8];int score[4];}stu[N];main(){input(stu);print(stu);}input(stu)struct student stu[];{int i,j;for(i=0;i<N;i++){printf("number");scanf("%s",stu.num);printf("name");scanf("%s",);for(j=0;j<3;j++){printf("\nscore\n");scanf("%d",&stu.score[j]);}printf("\n");}}print(stu)struct student stu[];{int i,j;printf("\nnumber name score1 score2 score3 \n"); for(i=0;i<N;i++){printf("%8s%10s",stu.num,);for(j=0;j<3;j++)printf("%7d",stu.score[j]);printf("\n");}}11.5struct student{char num[6];char name[8];int score[4];float avr;}stu[5];main(){int i,j,max,maxi,sum;float average;for(i=0;i<5;i++){printf("number");scanf("%s",stu.num);printf("name");scanf("%s",);for(j=0;j<3;j++){printf("\nscore\n");scanf("%d",&stu.score[j]);}}average=0;max=0;maxi=0;for(i=0;i<5;i++){sum=0;for(j=0;j<3;j++)sum+=stu.score[j];stu.avr=sum/3.0;average+=stu.avr;if(sum>max){max=sum;maxi=i;}}average/=5;printf("number name score1 score2 score3 average\n"); for(i=0;i<5;i++){printf("%8s%10s",stu.num,);for(j=0;j<3;j++)printf("%7d",stu.score[j]);printf("%6.2f\n",stu.avr);}printf("average=%5.2f\n",average);printf("The best student is %s,sum=%d\n",stu[maxi].name,max); }。

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