如何在C中打印内存地址 [英] How to printf a memory address in C

问题描述

我的代码是:

#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
", A,A,A);
    printf("[B] Dec: %d, Hex: %x, Unsigned: %u
", B,B,B);
    printf("[field width on B] 3: '%3u', 10: '%10u', '%08u'
", B,B,B);

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

我在 linux mint 中使用终端进行编译，当我尝试使用 gcc 进行编译时，我收到以下错误消息:

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
", &A);

我要做的就是打印变量 A 在内存中的地址.

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

推荐答案

使用格式说明符 %p:

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

<小时>

对于%p 说明符，标准要求参数的类型为void*.因为 printf 是一个可变参数函数，所以没有从 T * 到 void * 的隐式转换，这对于任何非可变参数函数在C. 因此，需要演员阵容.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/输出函数(C11 草案)

p 参数应该是一个指向 void 的指针.指针的值为转换为一系列打印字符，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.

而您使用的是 %x，它需要 unsigned int 而 &A 是 int * 类型>.您可以阅读手册中printf的格式说明符.printf 中的格式说明符不匹配导致未定义行为.

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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