这个双线戏法到底是怎么做到的呢？

至少有一些 C预处理器允许您将宏的值(而不是它的名称)字符串化，方法是将它通过一个类似函数的宏传递给另一个将其字符串化的宏:

#define STR1(x) #x
#define STR2(x) STR1(x)
#define THE_ANSWER 42
#define THE_ANSWER_STR STR2(THE_ANSWER) /* "42" */

示例用例 给你。

这确实有效，至少在 GCC 和 Clang (都是 -std=c99)中是这样，但我不确定 怎么做是否在 C 标准术语中有效。

这种行为是由 C99保证的吗？
如果是这样，C99如何保证它？
如果没有，那么在什么时候行为会从 C 定义变成 GCC 定义？

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最佳答案

Yes, it's guaranteed.

It works because arguments to macros are themselves macro-expanded, except where the macro argument name appears in the macro body with the stringifier # or the token-paster ##.

6.10.3.1/1:

... After the arguments for the invocation of a function-like macro have been identified, argument substitution takes place. A parameter in the replacement list, unless preceded by a # or ## preprocessing token or followed by a ## preprocessing token (see below), is replaced by the corresponding argument after all macros contained therein have been expanded...

So, if you do STR1(THE_ANSWER) then you get "THE_ANSWER", because the argument of STR1 is not macro-expanded. However, the argument of STR2 is macro-expanded when it's substituted into the definition of STR2, which therefore gives STR1 an argument of 42, with the result of "42".

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As Steve notes, this is guarenteed, and it has been guarenteed since the C89 standard -- that was the standard the codified the # and ## operators in macros and mandates recursively expanding macros in args before substituting them into the body if and only if the body does not apply a # or ## to the argument. C99 is unchanged from C89 in this respect.


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