问题
At least some C preprocessors let you stringize the value of a macro, rather than its name, by passing it through one function-like macro to another that stringizes it:
#define STR1(x) #x
#define STR2(x) STR1(x)
#define THE_ANSWER 42
#define THE_ANSWER_STR STR2(THE_ANSWER) /* "42" */
Example use cases here.
This does work, at least in GCC and Clang (both with -std=c99), but I'm not sure how it works in C-standard terms.
Is this behavior guaranteed by C99?
If so, how does C99 guarantee it?
If not, at what point does the behavior go from C-defined to GCC-defined?
回答1:
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".
回答2:
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.
来源：https://stackoverflow.com/questions/2751870/how-exactly-does-the-double-stringize-trick-work