typeof(func!0) != typeof(func!0())

Andrey Zherikov andrey.zherikov at gmail.com
Mon Aug 22 11:22:54 UTC 2022

On Monday, 22 August 2022 at 05:25:50 UTC, Ali Çehreli wrote:
> This is where the @property keyword makes a difference:
>     @property auto ref func(int i)() { return this; }
> Now U().func!0 will be a call in your expression.

Is original `U().func!0` not a call?

> But I think std.traits.ReturnType is more explicit and does 
> work in this case:
>         import std.traits;
>         alias type = ReturnType!(U().func!0);

This works but looks strange - I'm checking the type of UDA 
@(U().func!0) int b;

pragma(msg, __traits(getAttributes, b));             		// 
pragma(msg, typeof(__traits(getAttributes, b)[0]));  		// pure 
nothrow @nogc ref @safe U() return
pragma(msg, ReturnType!(__traits(getAttributes, b)[0]));  	// U
pragma(msg, is(typeof(__traits(getAttributes, b)[0]) : U)); // 
pragma(msg, hasUDA!(b, U));                          		// false

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