<code>
void f(...){}
void f(int i, ...){}
void f(int i, int j){}
void main(){	
    f(0, 0);	
}
</code>

The call "f(0,0)" matches all three definitions of function f, which should not compile.

On Tue, 20 Sep 2005 14:20:52 +0800, zwang wrote:

> <code>
> void f(...){}
> void f(int i, ...){}
> void f(int i, int j){}
> void main(){
>      f(0, 0);
> }
> </code>
> 
> The call "f(0,0)" matches all three definitions of function f, which should not compile.

But using the rules for specialization, the f(0,0) is an exact match so it
compiles.

-- 
Derek
(skype: derek.j.parnell)
Melbourne, Australia
20/09/2005 4:48:28 PM

On Tue, 20 Sep 2005 14:20:52 +0800, zwang wrote:

> <code>
> void f(...){}
> void f(int i, ...){}
> void f(int i, int j){}
> void main(){
>      f(0, 0);
> }
> </code>
> 
> The call "f(0,0)" matches all three definitions of function f, which should not compile.

But a more problematic issue is that "f(0, 'a')" fails to compile.

test.d(5): function test.f called with argument types:
        (int,char)
matches both:
        test.f(...)
and:
        test.f(int,int)



And I can't work out how to get it to call 'f(int, ...)' if I want to.

-- 
Derek
(skype: derek.j.parnell)
Melbourne, Australia
20/09/2005 4:58:35 PM