ANSI CPP ## operator

[Jim Gardner]

In article <290 at alice.marlow.reuters.co.uk> fox at alice.marlow.reuters.co.uk (Paul Fox) writes:
>ANSI gets around this via the '#' and '##' operators -- available within
>the preprocessor phase only. As I read it 
>
...
># define	CTRL(x)		'##x##' & 037
>		CTRL(A);
>
>gives:		'A' & 037

## concatenates two preprocessor tokens. Is "'" guaranteed to be a valid
pre-processor symbol? I don't think so, so the result is implementation
defined. Similarly, if the result of ## is not a valid token, the result is
implementation defined (or some other term, I forget, which means unpredictable
results). So after the first "'##x" (or "x##'" if the implementation does it
that way), the result is not a valid token, and the implementation might do
something strange. An example of possible behaviour that I think conforms with
the standard:
	the "'" is an invalid token, and is left as a special internal token.
			(i.e., same token type generated by an @)
	"'##x" doesn't form a valid token, so it is left as two tokens
			(invalid token and the x macro arg)
	then a similar thing happens with "x##'", resulting in three
		separate tokens.
Another possibility is that the "'##x##'" is interpretted as a character
constant which might result in an error or warning about too many chars
in a char constant (actually, I think this is more likely, maybe even 
required by the ANSI standard). To see this in another light consider
#define f(a)  "##a##"
f(hi there)