More operators inside `is(...)` expressions

Timon Gehr timon.gehr at
Mon Aug 24 17:17:15 UTC 2020

On 24.08.20 13:49, Steven Schveighoffer wrote:
> Yes. If you write !is(T == void), then you are already not checking 
> whether T is defined. This is no different.

So far the pattern is `is(S op T)`. It checks whether S is a valid type 
and then checks `S op T`.

> This literally is just a nicer way to write it, where the operation is 
> closer to the parameters, instead of partly outside the expression.

(I think the implications any of this has for the compiler 
implementation or the user experience has been greatly exaggerated.)

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