让学指针变得更简单（三）_南航科院小张

网络投稿 02-07 2683

文章目录前言一、指针和数组笔试题解析1.1 一维数组1.2 字符数组1.3 二维数组2. 指针笔试题总结

前言

指针面试题，对指针的理解不再停留在简单的知识层面上，而是可以知道面试题中指针的考察是怎样的；

一、指针和数组笔试题解析 1.1 一维数组

首先说一下知识点：很重要！！！数组名的意义：

sizeof(数组名),这里的数组名表示整个数组,计算的是整个数组的大小。&数组名，这里的数组名表示整个数组，取出的是整个数组的地址。除此之外所有的数组名都表示首元素的地址。

int a[ ] = {1,2,3,4}; printf( “%d\n”,sizeof(a) ); printf( “%d\n”,sizeof(a+0) ); printf( “%d\n”,sizeof(a) ); printf( “%d\n”,sizeof(a+1) ); printf( “%d\n”,sizeof(a[1]) ); printf( “%d\n”,sizeof(&a) ); printf( “%d\n”,sizeof(&a) ; printf( “%d\n”,sizeof(&a+1) ); printf( “%d\n”,sizeof(&a[0]) ); printf( “%d\n”,sizeof(&a[0]+1) );

1.2 字符数组

char arr[] = {‘a’,‘b’,‘c’,‘d’,‘e’,‘f’}; printf("%d\n", sizeof(arr)); printf("%d\n", sizeof(arr+0)); printf("%d\n", sizeof(*arr)); printf("%d\n", sizeof(arr[1])); printf("%d\n", sizeof(&arr)); printf("%d\n", sizeof(&arr+1)); printf("%d\n", sizeof(&arr[0]+1)); printf("%d\n", strlen(arr)); printf("%d\n", strlen(arr+0)); printf("%d\n", strlen(*arr)); printf("%d\n", strlen(arr[1])); printf("%d\n", strlen(&arr)); printf("%d\n", strlen(&arr+1)); printf("%d\n", strlen(&arr[0]+1));

char arr[] = “abcdef”; printf("%d\n", sizeof(arr)); printf("%d\n", sizeof(arr+0)); printf("%d\n", sizeof(*arr)); printf("%d\n", sizeof(arr[1])); printf("%d\n", sizeof(&arr)); printf("%d\n", sizeof(&arr+1)); printf("%d\n", sizeof(&arr[0]+1)); printf("%d\n", strlen(arr)); printf("%d\n", strlen(arr+0)); printf("%d\n", strlen(*arr)); printf("%d\n", strlen(arr[1])); printf("%d\n", strlen(&arr)); printf("%d\n", strlen(&arr+1)); printf("%d\n", strlen(&arr[0]+1));

char *p = “abcdef”; printf("%d\n", sizeof§); printf("%d\n", sizeof(p+1)); printf("%d\n", sizeof(*p)); printf("%d\n", sizeof(p[0])); printf("%d\n", sizeof(&p)); printf("%d\n", sizeof(&p+1)); printf("%d\n", sizeof(&p[0]+1)); printf("%d\n", strlen§); printf("%d\n", strlen(p+1)); printf("%d\n", strlen(*p)); printf("%d\n", strlen(p[0])); printf("%d\n", strlen(&p)); printf("%d\n", strlen(&p+1)); printf("%d\n", strlen(&p[0]+1));

1.3 二维数组

int a[3][4] = {0}; printf( “%d\n”,sizeof(a) ); printf( “%d\n",sizeof(a[0][0]) ); printf( “%d\n”,sizeof(a[0]) ); printf( ”%d\n",sizeof(a[0]+1) ); printf( “%d\n”,sizeof(* (a[0]+1)) ); printf( “%d\n”,sizeof(a+1) ); printf( “%d\n”,sizeof(* ( a+1) ) ); printf( “%d\n”,sizeof(&a[0]+1) ); printf( “%d\n”,sizeof( * (&a[0]+1) ) ); printf( “%d\n”,sizeof(*a) ); printf( “%d\n”,sizeof(a[3]) );

2. 指针笔试题

int main() { int a[5] = { 1, 2, 3, 4, 5 }; int *ptr = (int *)(&a + 1); printf( “%d,%d”, *(a + 1), *(ptr - 1)); return 0; } //程序的结果是什么？

图片上的 52是5哈，多了个2，不好意思😣

//由于还没学习结构体，这里告知结构体的大小是20个字节 struct Test { int Num; char *pcName; short sDate; char cha[2]; short sBa[4]; }*p; //假设p 的值为0x100000。如下表表达式的值分别为多少？ //已知，结构体Test类型的变量大小是20个字节 int main() { printf("%p\n", p + 0x1); printf("%p\n", ( unsigned long )p + 0x1); printf("%p\n", (unsigned int * )p + 0x1); return 0; }

int main() { int a[4] = { 1, 2, 3, 4 }; int *ptr1 = (int *)(&a + 1); int *ptr2 = (int *)((int)a + 1); printf( “%x,%x”, ptr1[-1], *ptr2); return 0; }

#include <stdio.h> int main() { int a[3][2] = { (0, 1), (2, 3), (4, 5) }; int *p; p = a[0]; printf( “%d”, p[0]); return 0; }

int main() { int a[5][5]; int(*p)[4]; p = a; printf( “%p,%d\n”, &p[4][2] - &a[4][2], &p[4][2] - &a[4][2]); return 0; }

int main() { int aa[2][5] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int *ptr1 = (int *)(&aa + 1); int *ptr2 = (int * )( * (aa + 1) ); printf( “%d,%d”, * (ptr1 - 1), *(ptr2 - 1) ); return 0; }

#include <stdio.h> int main() { char *a[] = {“work”,“at”,“alibaba”}; char**pa = a; pa++; printf("%s\n", *pa); return 0; }

int main() { char * c[ ] = {“ENTER”,“NEW”,“POINT”,“FIRST”}; char ** cp[] = {c+3,c+2,c+1,c}; char *** cpp = cp; printf("%s\n", ** ++cpp); printf("%s\n", * --* ++cpp+3); printf("%s\n", * cpp[-2]+3); printf("%s\n", cpp[-1][-1]+1); return 0; }

总结

指针的博客就告一段落了，敬请期待后面的博客吧！！！