如何给printf在C中的内存地址 [英] How to printf a memory address in C

问题描述

我的code是：

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

void main()
    {
    char string[10];
    int A = -73;
    unsigned int B = 31337;

    strcpy(string, "sample");

    // printing with different formats
    printf("[A] Dec: %d, Hex: %x, Unsigned: %u\n", A,A,A);
    printf("[B] Dec: %d, Hex: %x, Unsigned: %u\n", B,B,B);
    printf("[field width on B] 3: '%3u', 10: '%10u', '%08u'\n", B,B,B);

    // Example of unary address operator (dereferencing) and a %x
    // format string 
    printf("variable A is at address: %08x\n", &A);

我使用的终端Linux Mint的编译，当我尝试用gcc我收到以下错误信息进行编译：

I am using the terminal in linux mint to compile, and when I try to compile using gcc I get the following error message:

basicStringFormatting.c: In function ‘main’:
basicStringFormatting.c:18:2: warning: format ‘%x’ expects argument
of type ‘unsigned int’, but argument 2 has type ‘int *’ [-Wformat=]
printf("variable A is at address: %08x\n", &A);

所有我想要做的就是在变量A的打印内存中的地址。

All I am trying to do is print the address in memory of the variable A.

推荐答案

使用格式说明％P ：

printf("variable A is at address: %p\n", (void*)&A);

---

该标准要求的参数类型是无效* 为％P 说明符。一直以来，的printf 是一个可变参数函数，没有隐式转换无效* 从 T * 这将隐含发生在C.任何非可变参数的功能因此，需要投。引用标准：

---

The standard requires that the argument is of type void* for %p specifier. Since, printf is a variadic function, there's no implicit conversion to void * from T * which would happen implicitly for any non-variadic functions in C. Hence, the cast is required. To quote the standard:

7.21.6格式化输入/输出功能（C11草案）

P中的参数应该是一个指针无效。指针的值是
  转换为打印字符的序列，在一个
  实现定义的方式。

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.

而你正在使用％X ，它期望 unsigned int类型，而＆安培; A 的类型为为int * 。你可以阅读从手动格式说明的printf。在printf的格式说明不匹配导致的未定义行为的。

Whereas you are using %x, which expects unsigned int whereas &A is of type int *. You can read about format specifiers for printf from the manual. Format specifier mismatch in printf leads to undefined behaviour.

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