如何在C中打印变量地址? [英] How to print variable addresses in C?

问题描述

当我运行此代码时.

#include <stdio.h>

void moo(int a, int *b);

int main()
{
    int x;
    int *y;

    x = 1;
    y = &x;

    printf("Address of x = %d, value of x = %d
", &x, x);
    printf("Address of y = &d, value of y = %d, value of *y = %d
", &y, y, *y);
    moo(9, y);
}

void moo(int a, int *b)
{
    printf("Address of a = %d, value of a = %d
", &a, a);
    printf("Address of b = %d, value of b = %d, value of *b = %d
", &b, b, *b);
}

我的编译器不断收到这个错误.

I keep getting this error in my compiler.

/Volumes/MY USB/C Programming/Practice/addresses.c:16: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
/Volumes/MY USB/C Programming/Practice/addresses.c:17: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int **’
/Volumes/MY USB/C Programming/Practice/addresses.c:17: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘int *’
/Volumes/MY USB/C Programming/Practice/addresses.c: In function ‘moo’:
/Volumes/MY USB/C Programming/Practice/addresses.c:23: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
/Volumes/MY USB/C Programming/Practice/addresses.c:24: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int **’
/Volumes/MY USB/C Programming/Practice/addresses.c:24: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘int *’

你能帮我吗?

谢谢

吹牛

推荐答案

您想使用 %p 打印一个指针.来自规范:

You want to use %p to print a pointer. From the spec:

p 参数应该是一个指向 void 的指针.指针的值以实现定义的方式转换为打印字符序列.

p The argument shall be a pointer to void. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.

不要忘记演员表，例如

printf("%p
",(void*)&a);

这篇关于如何在C中打印变量地址?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！