1. sizeof和strlen的对⽐

1.1 sizeof

在学习操作符的时候，我们学习了 sizeof ， sizeof 计算变量所占内存空间⼤⼩的，单位是字 节，如果操作数是类型的话，计算的是使⽤类型创建的变量所占内存空间的⼤⼩。 sizeof 只关注占⽤内存空间的⼤⼩，不在乎内存中存放什么数据。

⽐如：

#include <stdio.h> int main() { int a = 10; printf("%d\n", sizeof(a)); printf("%d\n", sizeof a); printf("%d\n", sizeof(int)); return 0; }

1.2 strlen

strlen 是C语⾔库函数，功能是求字符串⻓度。函数原型如下：

size_t strlen ( const char * str );

统计的是从 strlen 函数的参数 str 中这个地址开始向后， \0 之前字符串中字符的个数。 strlen 函数会⼀直向后找 \0 字符，直到找到为⽌，所以可能存在越界查找。

#include <stdio.h> int main() { char arr1[3] = {'a', 'b', 'c'}; char arr2[] = "abc"; printf("%d\n", strlen(arr1)); printf("%d\n", strlen(arr2)); printf("%d\n", sizeof(arr1)); printf("%d\n", sizeof(arr2)); return 0; }

1.3 sizeof和strlen的对⽐

2. 数组和指针笔试题解析

数组名的意义：

1. 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。

2. &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。

3. 除此之外所有的数组名都表⽰⾸元素的地址。

2.1 ⼀维数组

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));

2.2 字符数组

代码1：

#include <stdio.h> int main() { 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)); return 0; }

代码2：

#include <stdio.h> #include <string.h> int main() { char arr[] = {'a','b','c','d','e','f'}; 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)); return 0; }

代码3：

#include <stdio.h> int main() { 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)); return 0; }

代码4：

#include <stdio.h> #include <string.h> int main() { char arr[] = "abcdef"; 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)); return 0; }

代码5：

#include <stdio.h> int main() { char *p = "abcdef"; printf("%d\n", sizeof(p)); 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)); return 0; }

代码6：

#include <stdio.h> #include <string.h> int main() { char *p = "abcdef"; printf("%d\n", strlen(p)); 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)); return 0; }

2.3 ⼆维数组

#include <stdio.h> int main() { 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])); return 0; }

3. 指针运算笔试题解析

3.1 题⽬1：

#include <stdio.h> 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？

3.2 题⽬2

//在X86环境下 //假设结构体的⼤⼩是20个字节 //程序输出的结果是啥？ struct Test { int Num; char *pcName; short sDate; char cha[2]; short sBa[4]; }*p = (struct Test*)0x100000; int main() { printf("%p\n", p + 0x1); printf("%p\n", (unsigned long)p + 0x1); printf("%p\n", (unsigned int*)p + 0x1); return 0; }

3.3 题⽬3

#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; }

3.4 题⽬4

//假设环境是x86环境，程序输出的结果是啥？ #include <stdio.h> 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; }

3.5 题⽬5

#include <stdio.h> 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; }

3.7 题⽬7

#include <stdio.h> 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; }