Re: [Dev-C++] printf

[Esteban V. <est...@ho...>]

That's mean that "(void _ss *) _va_ptr" is the way to obtain the variable
list of arguments ? For example

myprintf("%d%c%f", 1, 'a', 1.23);

the variable list (1, 'a' and 1.23) is passed to the printf inside the
function through (void _ss *) _va_ptr ?

bye


----- Original Message -----
From: "Goran Kavrecic" <go...@vo...>
To: "DevC++" <dev...@li...>
Sent: Monday, 27 January, 2003 13:03
Subject: Re: [Dev-C++] printf


Here you will find a sample of something i've done for a proprietary display

*.h:
int printfg( const char *fmt, ... );

*.cpp:
int printfg( const char *fmt, ... )
{
    char out[200];
      int i = vsprintf( out,fmt,(void _ss *) _va_ptr );

   // out is formated as would be called sprintf
   int x=getx();
   int y=gety();
   int dx=8*strlen(out)-1;
   int dy=8;
   dispmy( out,x+8*i,y );
   return i;
}


----- Original Message -----
From: "Esteban Valverde" <est...@ho...>
To: "dev-cpp" <dev...@li...>
Sent: Monday, January 27, 2003 4:50 PM
Subject: [Dev-C++] printf


> Hi, I wish to overload the printf function to do the same thing plus
others,
> just like...
>
> void myprintf ( const char *format, ... )
> {
>     //   I do my work here
>
>     printf( format, ... );
> }
>
> How must I do to pass the variable argument list (...) from the definition
> of myprintf to the printf function ?
>
> Bye, Esteban
>
>
>
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