c char到int指针

Question

#include<stdio.h>

int main(void)    
{
    int arr[3] = {2, 3, 4};
    char *p;
    p = (char*)arr;
    printf("%d", *(int*)(p+1));
    return 0;
}

我在指针上做了这个问题。 自从char *转换为int *以来，我期望输出为某些垃圾值，但输出始终为50331648，这很奇怪。 请解释

编辑：我在某些网站上阅读了此输出问题，因此需要根据给定的说明进行输出

Answer 1

您正在执行的操作是未定义的行为。

假设sizeof(int)为4

小端系统

在小端系统中， arr的内存布局为：

arr
+----+----+----+----+----+----+----+----+----+----+----+----+
| 02 | 00 | 00 | 00 | 03 | 00 | 00 | 00 | 04 | 00 | 00 | 00 |
+----+----+----+----+----+----+----+----+----+----+----+----+

使用时：

char* p = arr;

p    p+1
|    |
v    v
+----+----+----+----+----+----+----+----+----+----+----+----+
| 02 | 00 | 00 | 00 | 03 | 00 | 00 | 00 | 04 | 00 | 00 | 00 |
+----+----+----+----+----+----+----+----+----+----+----+----+

如果将p+1解释为int* ，并将该位置处的对象评估为int ，则会得到：

+----+----+----+----+
| 00 | 00 | 00 | 03 |
+----+----+----+----+

在小端系统中，该数字为

0x03000000

等于50331648 ，这是您获得的输出。

大端系统

在大端系统中， arr的内存布局为：

arr
+----+----+----+----+----+----+----+----+----+----+----+----+
| 00 | 00 | 00 | 02 | 00 | 00 | 00 | 03 | 00 | 00 | 00 | 04 |
+----+----+----+----+----+----+----+----+----+----+----+----+

评估*(int*)(p+1)时得到的数字是：

+----+----+----+----+
| 00 | 00 | 02 | 00 |
+----+----+----+----+

等于0x0200 ，即512。

很明显，为什么您执行的操作是未定义的行为。

Answer 2

您将arr声明为int，将p声明为char 。 您的程序将无法使用现代编译器进行编译。 您尝试在其中分配不同类型的指针，这种方法是错误的。

我认为这是您需要的：

#include<stdio.h>

int main(void)    {
    int arr[3] = {2, 3, 4};
    char *p;
    p = (char*)arr;
    printf("%d", *(int*)(p+1));
    return 0;
}

输出（垃圾）：

50331648

Answer 3

这是您的代码的逐行说明，最后说明了为什么获得该输出

//assuming 32-bit machine 
int arr[3] = {2, 3, 4}; //arr is a pointer to the first element which is an int 2 with a memory size of 4 bytes (int = 4 bytes)
char *p; //p is declared a pointer to a char implying it points to a 1 byte memory size (char = 1 byte)
p = (char*)arr; /* p now points to the first element of the array but actually this happens by luck. 
                  here is why: (char*)arr will attempt to cast the int pointer to a char pointer, 
                  in other words, asking it to present whatever took 4bytes to be presented in 1 byte 
                  which will overwrite the least significant bytes. but since numbers are stored in reverse order, 
                  it will overwrite only zeros causing no problem.*/
printf("%d", *(int*)(p+1)); /* p is still a char pointer holding memory address of 2 in the array. 
                            casting that to int* will add extra bytes from memory. so finally what you're getting is 
                            garbage number which I think is the virtual memory limit, (48 Gb *1024 *1024 = 50331648 Kb)*/