在C中打印指针

我试图用指针理解一些东西,所以我写了这段代码:

#include 

int main(void)
{
    char s[] = "asd";
    char **p = &s;

    printf("The value of s is: %p\n", s);
    printf("The direction of s is: %p\n", &s);

    printf("The value of p is: %p\n", p);
    printf("The direction of p is: %p\n", &p);

    printf("The direction of s[0] is: %p\n", &s[0]);
    printf("The direction of s[1] is: %p\n", &s[1]);
    printf("The direction of s[2] is: %p\n", &s[2]);

    return 0;
}

用gcc编译时我得到以下警告:

$ gcc main.c -o main-bin -ansi -pedantic -Wall -lm
main.c: In function ‘main’:
main.c:6: warning: initialization from incompatible pointer type
main.c:9: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char (*)[4]’
main.c:11: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char **’
main.c:12: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char ***’

(gcc的标志是因​​为我必须是C89)

为什么不兼容的指针类型？数组的名称不是指向它的第一个元素的指针吗？所以,如果s是指向'a'的指针,那么&s必须是a char **,不是吗？为什么我会收到其他警告？我是否必须使用(void *)转换指针才能打印它们？

跑步的时候我得到这样的东西:

$ ./main-bin
The value of s is: 0xbfb7c860
The direction of s is: 0xbfb7c860
The value of p is: 0xbfb7c860
The direction of p is: 0xbfb7c85c
The direction of s[0] is: 0xbfb7c860
The direction of s[1] is: 0xbfb7c861
The direction of s[2] is: 0xbfb7c862

s的价值和它的方向(当然还有价值p)如何相同？

"s"不是"char*",它是"char [4]".所以,"&s"不是"char**",而是"指向4个字符阵列的指针".您的编译器可能会将"&s"视为您编写了"&s [0]",这大致相同,但它是"char*".

当你写"char**p =&s;" 你试图说"我希望将p设置为当前指向"asd"的东西的地址.但是目前没有任何东西指向 "asd".只有一个数组保存 "asd";

char s[] = "asd";
char *p = &s[0];  // alternately you could use the shorthand char*p = s;
char **pp = &p;

是的,您的编译器期望void*.只是把它们变成无效*.

/* for instance... */
printf("The value of s is: %p\n", (void *) s);
printf("The direction of s is: %p\n", (void *) &s);