从（char *）到（int *）取消引用的指针在此示例中未被理解

Question

我正在网站上对C进行练习测试，我碰巧看到了这个问题。 我的疑问在评论中有解释，所以请阅读它们。

#include<stdio.h>

int main()
{
    int arr[3] = {2, 3, 4};  // its assumed to be stored in little-endian i.e;
                             // 2 = 00000010 00000000 00000000 00000000
                             // 3 = 00000011 00000000 00000000 00000000
                             // 4 = 00000100 00000000 00000000 00000000   

    char *p;

    p = arr;

    p = (char*)((int*)(p));  

    printf("%d ", *p);   

    p = (int*)(p+1);       // This casting is expected to convert char pointer p 
                           // to an int pointer , thus value at p ,now is assumed 
                           // to be equal to 00000000 00000000 00000000 00000011 
                           // but, the output was : 0  . As ,per my assumption it
                           // should be : 2^24+2^25 = 50331648 ,Please Clarify 
                           // if my assumption is Wrong and explain Why?

    printf("%d\n", *p);

    return 0;
 }

Answer 1

如果您要将p投回int*，那么int值将为：

00000000 00000000 00000000 00000011

其中最后一个字节是第二个数组元素的第一个字节。通过执行p+1，您将跳过第一个元素的最不重要的字节。

请记住，p仍然是一个字符指针，因此为其分配int*不会改变它的类型。

当printf p+1处的字符时，您正在打印第二个字节的值，即0。

Answer 2

请记住p仍然是char - 指针。因此*p从中获取char值。当char值作为参数传递给可变参数函数（如int）时，printf值将被提升为{{1}}。

Answer 3

p = (char*)((int*)(p));
// till now the pointer p is type casted to store the variable of type character.

printf("%d, ", *p); // %d means integer value so value at first address i.e. 2 will be printed.

p = (int*)(p+1); // here p is still of type character as type casted in step 1 so p(i.e address) and plus 1 will increase only by one byte so  

假设整数需要2个字节的存储空间 整数数组将作为

存储在内存中

value 2 3 4
address 00000010 00000000 00000011 00000000 00000100 00000000 
pointer p+1 

所以p+1指向未填充的位置，因为初始化期间2,3,4存储在整数类型的变量（2个字节）中。

因此p+1将指向00000000。

(int*)p+1 // p+1 is type casted again to integer

printf("%d", *p); // this will print 0 as output as by default integer contains 0 as value.