c - 通过*指针分配后，相邻的内存块将被零填充。为什么？

Question

我正在通过一本旧书“C 编程语言”学习 C，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;
}

所以，输出是:

AAAA
0000

为什么最后 4 个字节为零？

附注已经得到答案:类型 long 在 x64 机器上是 8 个字节，而我不善于观察。惊讶 StackOverflow 是多么好的东西。谢谢你们。

Answer 1

您的 long 可能是 64 位大。它可以与 int32_t 指针一起使用(在我的电脑上):

#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，这种类型双关是严格别名违规(这使得您的程序未定义)。 memcpy 或 char 逐个char 从 32 位整数复制，或 unions (最安全，以防万一决定开始动态分配对象)，应该可靠地做到这一点:

#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;
}