Question

我试图通过自己澄清指针指向的地址，指针本身的地址和地址所指的值来更好地理解简单指针。所以我写了一小段代码：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int a;
    int *p;
    int **pp;

    a = 42;

    /* Take the address of a */
    p = &a;

    /* Take the address of p */
    pp = &p;

    printf("Address of int &a:                          %p\n\n", &a);
    printf("value of a:                                 %d\n\n", a);

    printf("Address where *p points to via (void *)p:   %p\n\n", (void *)p);
    printf("Value that *p points to via *p:             %d\n\n", *p);
    printf("Address of *p itself via (void *)&p:        %p\n\n", (void *)&p);

    printf("Address where **p points to via (void *)pp: %p\n\n", (void *)pp);
    printf("Value that **pp points to via **pp:         %d\n\n", **pp);
    printf("Address of **p itself via (void *)&pp:      %p\n\n", (void *)&pp);
    return EXIT_SUCCESS;
}

这一切都按预期工作（如果我在这里犯了任何错误，请纠正我。）现在，我想更深入一级，并使用指向指针***ppp的指针，并为指针pp指向指向地址*p指向的地址。这是a的地址。以下是我认为我能做到的事情：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int a;
    int *p;
    int **pp;
    int **ppp;

    a = 42;

    /* Take the address of a */
    p = &a;

    /* Take the address of p */
    pp = &p;

    ppp = &pp;

    printf("Address of int &a:                               %p\n\n", &a);
    printf("value of a:                                      %d\n\n", a);

    printf("Address where *p points to via (void *)p:        %p\n\n", (void *)p);
    printf("Value that *p points to via *p:                  %d\n\n", *p);
    printf("Address of *p itself via (void *)&p:             %p\n\n", (void *)&p);

    printf("Address where **pp points to via (void *)pp:     %p\n\n", (void *)pp);
    printf("Value that **pp points to via **pp:              %d\n\n", **pp);
    printf("Address of **pp itself via (void *)&pp:          %p\n\n", (void *)&pp);

    printf("Address where ***ppp points to via (void *)ppp:  %p\n\n", (void *)ppp); 
    printf("Value that ***ppp points to via ***ppp:          %d\n\n", ***ppp);
    printf("Address where ***ppp points to via (void *)&ppp: %p\n\n", (void *)&ppp);

    return EXIT_SUCCESS;
}

但是这给了我一个不兼容的指针警告。有人可以向我解释为什么这不起作用以及对printf()的呼叫是否正确？

Answer 1

错误就在这一行 int ** ppp; / *它应该是*** ppp，因为你指向指向另一个的指针* /

Answer 2

问题在于：

int **ppp;

那应该是

int ***ppp;

现在你正试图强制一个三指针，成为一个双指针。这可能会在这一行上给你一个错误：

ppp = &pp;

您可能希望查看this article的第2位：）。

<强> P.S。如果您提供行号和特定错误，以备将来参考。人们可能想要更好，更好地帮助你：）

声明指向指针的指针

2 个答案: