preprocessor/9764: Varargs macro extension incorrectly expands if the varargs argument is the macro itself

[Neil Booth]

Ryan Mallon wrote:-

> #define func(a, varargs...) _func(a, ##varargs)
> 
> The problems occurs when the macro name itself is used in the varargs 
> part: e.g
> 
> func(a, func(b, c));
> 
> Incorrectly expands to:
> 
> _func(a, func(b, c));
> 
> Without the token paste opperator in the defintion it works as expected, 
> but then I cannot have varargs=0. Im working with a large amount of 
> existing code, so I cannot alter the actual calls on the macro definition.

What makes you think it's incorrect?

Neil.