Re: Will uniform function call syntax apply to operator overloads?

[Simen kjaeraas]

Peter Alexander <peter.alexander...@gmail.com> wrote:

In short, when UFC is working on all types, will this be possible:

Foo opBinary(string op)(Foo a, Foo b)
{
     return ...;
}

Foo x, y;
Foo z = x + y;

My reasoning here is that x + y is supposedly sugar for x.opBinary!("+")(y), so the free opBinary defined above could be chosen as a pseudo member of Foo.

Will this be possible?

As long as operator overloading is defined the way it is, it should work
like that, yes.


--
Simen