目录

1. sizeof和strlen的对⽐

1.1 sizeo

1.2 strlen

1.3 sizeof和strlen的对⽐

2. 数组和指针笔试题解析

2.1 ⼀维数组

2.2 字符数组

2.3 ⼆维数组

3. 指针运算笔试题解析

3.1 题⽬1：

3.2 题⽬2

3.3 题⽬3

3.4 题⽬4

3.5 题⽬5

3.6 题⽬6

3.7 题⽬7

---

1. sizeof和strlen的对⽐

1.1 sizeo

        在学习操作符的时候，我们学习了 sizeof ， sizeof 计算变量所占内存内存空间⼤⼩的，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：

#inculde <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：

1 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

sizeof	strlen
sizeof是操作符	strlen是库函数，使⽤需要包含头⽂件 string.h
sizeof计算操作数所占内存的 ⼤⼩，单位是字节	srtlen是求字符串⻓度的，统计的是 \0 之前字符的隔个数
不关注内存中存放什么数据	关注内存中是否有 \0 ，如果没有 \0 ，就会持续往后找，可能 会越界

2. 数组和指针笔试题解析

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：

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));

代码2： 

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));

代码3： 

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));

代码4： 

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));

代码5： 

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));

代码6： 

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));

2.3 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]));

        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。

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;
}

3.6 题⽬6

#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;
}