Question on ANSI ## pre-processor operator.

[Stephen Clamage]

knurlin at spf.trw.com (Scott Karlin) writes:

>I am looking for a clarification on the ANSI-C token merging
>operator (##).  According to H&S:  "After *all* [emphasis mine]
>macro replacements have been done, the two tokens surrounding
>any ## operator are combined into a single token."  I interpret
>this to mean that:

>#define INDEX 0
>#define FN_NAME(x) name ## x
>void FN_NAME(INDEX) () { printf("Hello\n"); }

>Should expand to:

>void name ## 0 () { printf("Hello\n"); }

>And then:

>void name0 () { printf("Hello\n"); }

Not quite.  Let's review.  In the absence of any # or ## tokens in a
macro replacement string, each macro parameter appearing in the
replacement string is replaced by its fully-macro-expanded actual
argument.  Example
	#define INDEX 0
	#define FN(x) fn x
	FN(INDEX)
becomes
	fn 0
because the actual argument INDEX is fully expanded before it replaces
the formal parameter x.

When a macro parameter in the replacement string is preceded by # or
preceded or followed by ##, the rule is different.  The actual argument
replaces the formal argument without any expansion, the # or ## operator
is applied, then the resulting text is rescanned.  This allows you to
build up macro names for further replacement.
Example:
	#define INDEX 0
	#define FN(x) fn ## x
	FN(INDEX)
becomes
	fn ## INDEX
which becomes
	fnINDEX
which cannot be further expanded.

Your example will not work with the extra level of macro replacement.
It can be used like this:
	#define FN_NAME(x) name ## x
	void FN_NAME(0) () { printf("Hello\n"); }
which yields
	void name0 () { printf("Hello\n"); }
-- 

Steve Clamage, TauMetric Corp, steve at taumet.com