Re: How can I test at compile time whether T is an instance of an interface ?

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On Wednesday, 23 September 2020 at 18:37:45 UTC, wjoe wrote:
I have some similar functions:
void register(C: IFoo)()
{
  _insert!C();
}

void register(C)() if (behavesLikeFoo!C)
{
  _insert!C();
}

There are more overloads with parameters so I want to merge them

void register(C, ARGS...)(ARGS args) if (behavesLikeFoo!C || 
isInstanceOf!(C, IFoo))
{
  _insert!C(args);
}

I found a lot of information on how to do this at runtime but 
not at compile time.
std.traits: isInstanceOf doesn't work. Neither did anything I 
tried with typeid etc.

The behavesLikeFoo constraint works as expected but it accepts 
any class no matter whether or not it implements the interface.

A class at compile time is it's own static type, OOP polymorphism 
is a runtime feature not compile time. You have to write your own 
traits for specific objects to get them to relate to each other 
using static overloading.