__func__未在预处理输出中替换 [英] __func__ not replaced in preprocessed output

问题描述

我正在阅读有关C/C ++中的__FUNCTION__/__func__的信息(它们用于打印使用它们的函数的名称).我读过的每个地方都说它们是宏，在预处理时会被替换.因此，我通过查看使用命令gcc -E prog.c的预处理输出来对此进行了调查.但是我发现__func__和__FUNCTION__都没有被函数名称的预处理器替换.

I was reading about __FUNCTION__ / __func__ in C/C++ (they are used to print the name of the function in which they are used). Every place I read they said that these are macros and are replaced at preprocessing time. So, I investigated this by seeing the preprocessed output using command gcc -E prog.c. But I saw that neither __func__ nor __FUNCTION__ were replaced by the preprocessor by the function name.

那么，它是一个宏吗?如果没有，它是什么以及如何实现?

So, is it a macro? If not, what is it and how is it implemented?

编辑

甚至尝试过cpp prog.c.但仍然没有被替换.

Even tried cpp prog.c. But still not replaced.

还 __FILE __，__ LINE__和__FUNCTION__在C ++中的用法这篇文章说，它从不影响性能.请澄清.

Also __FILE__, __LINE__, and __FUNCTION__ usage in C++ this post says that it never affects performance. Please clarify.

推荐答案

它们被实现为魔术变量". 手册说:

They are implemented as (and are) "magic variables". The manual says:

GCC提供了三个魔术变量，它们以字符串形式保存当前函数的名称. 其中的第一个是__func__，它是C99标准的一部分:

GCC provides three magic variables that hold the name of the current function, as a string. The first of these is __func__, which is part of the C99 standard:

标识符__func__由翻译器隐式声明，就像立即 按照每个函数定义的开头括号，声明

The identifier __func__ is implicitly declared by the translator as if, immediately following the opening brace of each function definition, the declaration

static const char __func__[] = "function-name";

出现

，其中function-name是词法包围函数的名称.这个名字 是该函数的未经修饰的名称.

appeared, where function-name is the name of the lexically-enclosing function. This name is the unadorned name of the function.

它们通常不能实现为预处理器宏；预处理器不解析函数作用域.当然，可以使预处理器 足够了解语法，以了解函数的开始和结束位置，但是通常它们不在该级别上运行.

They typically cannot be implemented as preprocessor macros; the preprocessor doesn't parse function scopes. Of course a preprocessor could be made to understand enough of the syntax to know where functions begin and end, but typically they're not operating at that level.

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