c - 为什么整数在 printf 中格式化指针时需要一个符号，而字符数组在 C 中不需要？

Question

具体来说，当我想格式化所述整数的指针时。我正在学习 The Art of Exploitation，它在以下代码片段中给我留下了深刻的印象。

#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]) {
    int value = 5;
    char buffer_one[8], buffer_two[8];

    strcpy(buffer_one, "one"); /* Put "one" into buffer_one. */
    strcpy(buffer_two, "two"); /* Put "two" into buffer_two. */

    printf("[BEFORE] buffer_two is at %p and contains \'%s\'\n", buffer_two, buffer_two);
    printf("[BEFORE] buffer_one is at %p and contains \'%s\'\n", buffer_one, buffer_one);
    printf("[BEFORE] value is at %p and is %d (0x%08x)\n", &value, value, value);

    printf("\n[STRCPY] copying %d bytes into buffer_two\n\n", strlen(argv[1]));
    strcpy(buffer_two, argv[1]); /* Copy first argument into buffer_two. */

    printf("[AFTER] buffer_two is at %p and contains \'%s\'\n", buffer_two, buffer_two);
    printf("[AFTER] buffer_one is at %p and contains \'%s\'\n", buffer_one, buffer_one);
    printf("[AFTER] value is at %p and is %d (0x%08x)\n", &value, value, value);
}

Answer 1

虽然 buffer_one 是 char[8] 类型，但在某些情况下，例如当作为函数参数传递时，它衰减到指向数组第一个元素的指针。该指针将是 char* 类型。

由于strcpy的第一个参数应该是char*类型，所以编译成功。