Re: how to handle void arguments in generic programming ?

[Philippe Sigaud]

On Mon, Nov 11, 2013 at 4:52 AM, Timothee Cour <thelastmamm...@gmail.com> wrote:
>
> The code snippet below doesn't work. Is there a way to make it work?
>
> import std.stdio;
> void main(){
>   writelnIfNonVoid(writeln("ok"));
> }
> void writelnIfNonVoid(T...)(T a){
>   static if(T.length)
>     writeln(a);
> }

Shouldn't that be:

import std.stdio;
void main(){
  writelnIfNonVoid("ok"); // not writeln("ok")
  writelnIfNonVoid();
}
void writelnIfNonVoid(T...)(T a){
  static if(T.length)
    writeln(a);
}

or, if you want to print an expression whenever its type is not
`void`, then test it with an `is()`:

import std.stdio;
import std.traits: isCallable, ReturnType;

import std.stdio;
void main(){
  writelnIfNonVoid("ok");
  writelnIfNonVoid({ writeln("ok");}); //  { ... ;} is an anonymous function
}

void writelnIfNonVoid(T)(T a){
  static if(isCallable!a && !is(ReturnType!(a) == void))
    writeln(a()); // calling
  else static if(!isCallable!a && !is(T == void))
   writeln(a);
}