第十章 指针(含答案)

C语言指针习题和答案精选文档TTMS system office room 【TTMS16H-TTMS2A-TTMS8Q8-一、用指针方法编写一个程序，输入3个整数，将它们按由小到大的顺序输出#include <>void swap(int *pa,int *pb){int temp;temp = *pa;*pa = *pb;*pb = temp;}void main(){int a,b,c,temp;scanf("%d%d%d",&a,&b,&c);if(a>b)swap(&a,&b);if(b>c)swap(&b,&c);if(a>c)swap(&a,&c);printf("%d,%d,%d",a,b,c);}二、C语言用指针方法输入3个字符串按由小到大顺序输出#include ""#include ""int main(int argc, char* argv[]){char *t;char *p1=NULL,*p2=NULL,*p3=NULL;char ch1[20]={0},ch2[20]={0},ch3[20]={0};p1=ch1;p2=ch2;p3=ch3;printf("No1:");scanf("%s",p1);fflush(stdin);printf("No2:");scanf("%s",p2);fflush(stdin);printf("No3:");scanf("%s",p3);fflush(stdin);if(strcmp(p1,p2)>0){t=p1;p1=p2;p2=t;}if(strcmp(p1,p3)>0){t=p1;p1=p3;p3=t;}if(strcmp(p2,p3)>0){t=p2;p2=p3;p3=t;}printf("%s\n%s\n%s\n",p1,p2,p3);return 0;}编程输入一行文字，找出其中的大写字母，小写字母，空格，数字，及其他字符的个数#include<>void main(){int a=0,b=0,c=0,d=0,e=0,i=0;char *p,s[20];while((s[i]=getchar())!='\n')i++;p=s;while(*p!=10){if(*p>='A'&&*p<='Z')a++;else if(*p>='a'&&*p<='z')b++;else if(*p==' ')c++;else if(*p>='0'&&*p<='9')d++;else e++;p++;}printf("大写字母 %d 小写字母 %d\n",a,b);printf("空格 %d 数字 %d 非字符 %d\n",c,d,e);}写一个函数，将3 3矩阵转置#include ""void Transpose(int (*matrix)[3]){int temp;int i, j;for(i=1;i<3;i++)/*转置*/{for(j=0;j<i;j++){temp = *(*(matrix+j)+i);*(*(matrix+j)+i) = *(*(matrix+i)+j); *(*(matrix+i)+j) = temp;}}}void main(){int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}}; Transpose(a);for(int i = 0;i<3;i++){for(int j=0;j<3;j++){printf("%d ",a[i][j]);}printf("\n");}}}用指向一维数组的指针做函数参数#include<>#include<>int main(){void sort(char (*s)[6]);3”n",*(month_name+n));elseprintf("It is wrong.\n");}运行结果：① Input month: 2↙It is February.② Input month: 8↙It is August.③ Input month: 13↙It is wrong.9.11c语言：将字符串computer赋给一个字符数组，然后从第一个字母开始间隔地输出该串。

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;}。

1.若有说明:int i,j=7, *p=&i;,则与i=j;等价的语句是(B )。

(A)i= *p; (B)*p=*&j; (C)i=&j; (D)i=* *p;2.若有以下说明:int a[10]={1,2,3,4,5,6,7,8,9,10},*p=a;则数值为6的表达式是( C)。

(A)*p+6 (B)*(p+6) (C)*p+=5 (D)p+53.若有以下定义和语句:char *s1="12345",*s2="1234";printf("%d\n",strlen(strcpy(s1,s2)));则输出结果是( A)。

(A)4 (B)5 (C)9 (D)104.以下不能正确进行字符串赋初值的语句是(A )。

(A) char str[5]="good!"; (B) char str[]="good!";(C) char *str="good!"; (D) char str[5]={‘g',‘o',‘o',‘d'};5.若定义：int a=511,*b=&a;, 则printf("%d\n",*b);的输出结果为( D )。

(A) 无确定值 (B) a的地址 (C) 512 (D) 5116.下面程序的输出是( A )。

(A) 3 (B) 4 (C) 1 (D) 2main(){ int a[10]={ 1,2,3,4,5,6,7,8,9,10},*p=a;printf("%d\n",*(p+2));}7.若有说明：int n=2,*p=&n,*q=p;，则以下非法的赋值语句是（D ）。

(A) p=q; (B) *p=*q; (C) n=*q; (D) p=n;8.有如下说明int a[10]={1,2,3,4,5,6,7,8,9,10},*p=a;则数值为9的表达式是( B )。

第十章指针一．选择题1.变量的指针‎，其含义是指‎该变量的。

A）值B）地址C）名D）一个标志2.已有定义i‎n t k=2;int *ptr1,*ptr2;且ptr1‎和ptr2‎均已指向变‎量k，下面不能正‎确执行的赋‎值语句是。

A）k=*ptr1+*ptr2 B）ptr2=k C）ptr1=ptr2 D）k=*ptr1*(*ptr2) 3.若有说明：int *p,m=5,n;以下程序段‎正确的是。

A）p=&n ; B）p = &n ;scanf‎(“%d”,&p); scanf‎(“%d”,*p);C）scanf‎(“%d”,&n); D）p = &n ;*p=n ; *p = m ;4.已有变量定‎义和函数调‎用语句：int a=25;print‎_valu‎e(&a);下面函数的‎输出结果是‎。

void print‎_valu‎e(int *x){ print‎f(“%d\n”,++*x); }A）23 B）24 C）25 D）265.若有说明：int *p1, *p2,m=5,n;以下均是正‎确赋值语句‎的选项是。

A）p1=&m; p2=&p1 ; B）p1=&m; p2=&n; *p1=*p2 ;C）p1=&m; p2=p1 ; D）p1=&m; *p1=*p2 ;6.若有语句：int *p,a=4;和p=&a；下面均代表‎地址的一组‎选项是。

A）a,p,*&a B）&*a,&a,*p C）*&p,*p,&a D）&a,&*p,p7.下面判断正‎确的是。

A）char *a=”china‎”; 等价于char *a; *a=”china‎” ;B）char str[10]={“china‎”}; 等价于char str[10]; str[ ]={“china‎”;}C）char *s=”china‎”; 等价于char *s; s=”china‎” ;D）char c[4]=”abc”,d[4]=”abc”; 等价于char c[4]=d[4]=”abc” ;8.下面程序段‎中，for循环‎的执行次数‎是。

指针习题答案1．已知有以下的说明，int a[]={8,1,2,5,0,4,7,6,3,9};那么a[*(a+a[3])]的值为_ 0__.2．对于指向同一数组的指针变量，不能进行哪种运算? （A）A. +B. -C. =D. ==3．若有以下说明和语句，且0<=i<10, 则下面哪个是对数组元素的错误引用?（Ｃ）int a[]={1,2,3,4,5,6,7,8,9,0}, *p, i;p=a;A. *(a+i)B. a[p-a]C. p+iD. *(&a[i])4．若有以下说明和语句，且0<=i<10,则下面哪个是对数组元素地址的正确表示? （Ｄ）int a[]={1,2,3,4,5,6,7,8,9,0}, *p,i; p=a;A. &(a+1)B. a++C. &pD. &p[i]5．以下程序的输出结果是哪一项? （Ｄ）#includemain(){ int a[12]={1,2,3,4,5,6,7,8,9,10,11,12}, *p[4],i;for(i=0;i<4;i++)p[i]=&a[i*3];printf("%d\n",*p[3]);}A. 输出项不合法B. 4C. 7D. 106．有以下说明和语句，, 则p2-p1的值为多少?(单选) （A）int a[10], *p1, *p2;p1=a;p2=&a[5];A. 5B. 6C. 10D. 没有指针与指针的减法7．若有以下定义和语句:int s[4][5],(*ps)[5];ps=s;则对s数组元素的正确引用形式是（C）A) ps+1 B) *(ps+3) C) ps[0][2] D) *(ps+1)+38．若有以下的定义，int t[3][2]；能正确表示t数组元素地址的表达式是（C）A) &t[3][2]B) t[3]C) t[1]D) *t[2]9．以下程序的输出结果是a= 36 _, b= 16 _, c= 19 _。

《C++指针》习题参考答案(总9页)--本页仅作为文档封面，使用时请直接删除即可----内页可以根据需求调整合适字体及大小--《C++指针》习题学号姓名一、选择题1、有定义如下：int a[5], *p; 则下列描述错误的是 BA．表达式p=p+1是合法的B．表达式a=a+1是合法的C．表达式p-a是合法的D．表达式a+2是合法的2、函数定义为Fun(int &i),变量定义n=100,则下面调用该函数正确的是 CA．Fun(20) B．Fun(20+n) C．Fun(n)D．Fun(&n)3、若有说明int I,j=2,*p=&i;则能完成i=j赋值功能的语句是 BA．i=*p; B．*p=*&j; C．i=&j; D．i=**p;4、设已有定义char *st = "how are you";下列程序中正确的是 AA．char a[11],*p;strcpy(p=a+1,&st[4]);B．char a[11];strcpy(++a,st);C. char a[11];strcpy(a,st);D. char a[],*p;strcpy(p=a[1],st+2);5、设有数组int array[3][4]，下列引用数组元素的方法中错误的是 DA．array[i][j] B．*(*(array+i)+j)C. *(array[i]+j)D. *(array+i*4+j)6、有定义int s[4][5],(*ps)[5];ps=s;则对数组s中元素引用的正确的是 CA．ps+1 B．*(ps+3) C. ps[0][2] D. *(ps+1)+37、设有说明int b[4][4]，以下不能等价表示元素b[3][3]的是: BA．*&b[3][3] B．(*(*(b+3))+3) C．*(b[3]+3) D．*(*(b+3)+3)8、若有说明"char *s[]={"234","5678","9012","3456","7890"};",则表达式*s[1]> *s[3]比较的是 BA．"1234"和"9012" B．'5'和'3' C．'l'和'9' D．"5678"和"3456"9、设语句"char *s[]={"Student","Teacher","Father","Mother"},*ps=s[2];",执行语句"cout<<*s[1]<<','<<ps<<','<<*ps<<'\n';",则输出结果是 AA．T,Father,F B．Teacher,F,FatherC．Teacher,Father,Father D．语法错无输出10、设有以下说明语句：char sl[20]="hello",s2[20]="china";const char *ptr=s1;以下语句中不符合C++语法规则的语句是 DA．strcpy(s2,ptr); B．ptr=s2; C．ptr=ptr+4; D．*ptr=*s2;11、设s和t分别为指向两个长度相同的字符数组的指针，则循环语句while( C )；可以正确实现字符串t到s的完整复制功能。