On Friday, May 27, 2016 09:08:20 Marc Schütz via Digitalmars-d-learn wrote: > On Thursday, 26 May 2016 at 06:23:17 UTC, Jonathan M Davis wrote: > > The difference is that it's impossible to do > > 10.opBinary!"+"(15), so if you're forced to do > > foo.opBinary!"+"(bar) to get around a symbol conflict, it won't > > work with built-in types. > > Well, that begs the question: Why don't built-in types define > `opBinary`? That's just another arbitrary irregularity, isn't it.
It was never intended that any op* function be called by anyone except where the compiler lowers code to use them. They're for declaring overloaded operators on user-defined types so that those types can be used with those operators. If you're calling opBinary in your own code, you're doing it wrong. And it would be downright silly to then add opBinary to the built-in types. They don't need operator overloading. They already have the operators. Operators are supposed to be used as operators, not functions, and if there's any need to use them as functions, then there's something seriously wrong. And the fact that allowing free functions to overload operators via UFCS sends us into that territory just highlights the fact that they're a horrible idea. - Jonathan M Davis
