Simen kjaeraas napisał: > Peter Alexander <[email protected]> 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.
Funny. I remember asking this not too long ago and got no as an answer. :) Could someone from the D team take a stance on this? -- Tomek
