通过*指针分配后，相邻的内存空间将填充为零。为什么？

Question

我正在阅读一本旧书“ The C Programming Language”来学习C，并且目前正在尝试使用指针。

#include <stdio.h>
int
main (void)
{
    // init string
    char s[8] = "ZZZZZZZ";
    // it goes: Z Z Z Z Z Z Z \0

    long *p;         // make pointer refering to the same adress as s
    p = s;           // but declared long for modifying 4 bytes at once
    *p = 0x41414141; // and assign hexadecimal constant equal to 65 65 65 65

    // expect output to be: AAAAZZZ
    printf ("%s\n", s);
    // but get the next: AAAA

    // wrote the following line to find out what's wrong with last 4 chars
    printf ("%i%i%i%i\n", s[4], s[5], s[6], s[7]);
    // and those appear to become zero after messing with first 4 ones

    return 0;
}

因此，输出为：

AAAA
0000

为什么最后4个字节为零？

PS已经得到了答案：在x64机器上，long类型是8个字节，我很ob昧。惊讶StackOverflow是一件好事。感谢大伙们。

Answer 1

您的long大小可能为64位。它可以与int32_t指针一起使用（在我的PC上可以）：

#include <stdio.h>
#include <stdint.h>
int
main (void)
{
    // init string
    char s[8] = "ZZZZZZZ";
    // it goes: Z Z Z Z Z Z Z \0

    int32_t *p;         // making pointer refering to the same adress as s
    p = (int32_t*)s;           // but declared as long for modifying 4 bytes at once
    *p = 0x41414141; // and assign hexadecimal constant equal to 65 65 65 65

    // expect output to be: AAAAZZZ
    printf ("%s\n", s);
    // but get the next: AAAA

    // wrote the following line to find out what's wrong with last 4 chars
    printf ("%i%i%i%i\n", s[4], s[5], s[6], s[7]);
    // and those appear to become zero after messing with first 4 ones

    return 0;
}

但严格来说，这种类型修饰是一种严格的混叠冲突（这会使您的程序不确定）。memcpy或从32位整数进行char逐个char复制，或unions（如果您决定开始动态分配对象，这是最安全的），则应可靠地做到这一点：

#include <stdio.h>
#include <stdint.h>
#include <string.h>
int
main (void) 
{
    // init string
    char s[8] = "ZZZZZZZ";
    // it goes: Z Z Z Z Z Z Z \0

    int32_t src = 0x41414141;
    memcpy(s, &src, sizeof(src));

    // expect output to be: AAAAZZZ
    printf ("%s\n", s);
    // but get the next: AAAA

    // wrote the following line to find out what's wrong with last 4 chars
    printf ("%i%i%i%i\n", s[4], s[5], s[6], s[7]);
    // and those appear to become zero after messing with first 4 ones

    return 0;
}