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How, exactly, does the double-stringize trick work?

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?

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Peter Hosey Avatar asked May 01 '10 23:05

Peter Hosey


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1 Answers

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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Steve Jessop Avatar answered Sep 19 '22 01:09

Steve Jessop