Re: [cc65] function pointer, fastcall

Hi!

On Sat, Aug 16, 2008 at 01:30:10PM +0200, Stefan wrote:
> Ok, sorry i thought so because not all compilers support fastcall as a
> calling convention.
          
The problem wasn't the fastcall calling convention but the missing parenthesis
around the "*". You need a pointer to something, so you need to add "*". But
the parenthesis in a function bind tighter than the "*", so

        char *f(char);

is actually a function returning char*, not a pointer to a function returning
char. Using parens

        char (*f)(char);

you have a pointer to a function taking char and returning char. Now just add
fastcall

        char fastcall (*f)(char);

and you have a pointer to a fastcall function.
    
Regards


        Uz


-- 
Ullrich von Bassewitz                                  uz@musoftware.de
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