how to explain this

[zhangyefei.yefei@gmail.com]

#define cat(x,y) x##y concatenates x to y. But cat(cat(1,2),3) does
not expand but gives preprocessor warning. Why?

thanks .

[WANG CONG]

Re: how to explain this

zhangyefei.yefe i@gmail.com wrote:

<snip>

but i have another problem :

>But if you have...
>>
> #define xcat(x,y) x ## y
> #define cat(x,y) xcat(x,y)
>>
>...then...
>>
> cat(cat(1,2),3)
>>
>...proceeds as...
>>
> xcat(cat(1,2),3 )
>>
>...which is scanned for replacement. Note there's no ##
>in sight. The inner cat becomes...
>>
> xcat(xcat(1,2), 3)

>
why not xcat(1,2)##3 ?

Because the standard defines this.

Look at C99 6.10.3.1, it is said that:

"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 we can safely conclude that if arguments of a macro contains macros,
then the macros in arguments will be expanded first.

>
further more, macro replacement is how to implementation in c?
where i can find the source program.
thank you very much.

I think you can read the cpp source code in gcc, I guess it's right
in gcc/cppmacro.c .

Have fun!

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