函数的多重定义 [英] multiple definition of a function

问题描述

我定义了一个函数，当调试标志在头文件中关闭时显示一条消息，如下所示:

I defined a function to show a message when debug flags are off in a header file as below:

#ifdef  NDEBUG

#define debug_msg(expr, msg)        (static_cast<void>(0))

#else /* Not NDEBUG.  */

#ifndef SHOW_DEBUG_H_
#define SHOW_DEBUG_H_

#include <stdio.h>
void _show_in_debug(const char *_file, unsigned int _line,
        const char *_function, const char *_msg)
{
    printf("%s\t%d\t%s\t%s\n", _file, _line, _function, _msg);
    fflush(NULL);
}

#endif

#define debug_msg(expr, msg)                \
  ((expr)                               \
   ? _show_in_debug(__FILE__, __LINE__, __func__, msg)  \
   : static_cast<void>(0))

#endif

当我在文件中包含头文件时，出现以下错误:

when I include the header in more than a file, I get the following error:

_show_in_debug(char const *，unsigned int，charconst *，char const *)'

multiple definition of `_show_in_debug(char const*, unsigned int, char const*, char const*)'

我不完全知道自己在做什么错，有什么帮助吗?

I don't exactly know what I am doing wrong here, any help ?

推荐答案

即使有了包含保护，您最终在每个编译单元中的定义仍为 show_in_debug .然后，将这些单元链接起来会导致多定义错误.

Even with the include guards, you end up with a definition of _show_in_debug in each compilation unit. Linking those units then results to a multiple definition error.

对于这样的调试功能，请将功能定义为 static ，以使其在其编译单元之外不可见:

For a debugging function like this, define the function as static so that it is not visible outside its compilation unit:

static void _show_in_debug(const char *_file, unsigned int _line,
        const char *_function, const char *_msg)
{
    printf("%s\t%d\t%s\t%s\n", _file, _line, _function, _msg);
    fflush(NULL);
}

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