On Fri, Aug 01, 2014 at 02:20:53PM +0000, "Nordlöw" via Digitalmars-d-learn
wrote:
[...]
> What is the preffered (fast) way to check at compile-time if an
> instance x of a type T can be used *either* as
>
> f(x)
>
> or
>
> x.f?
if (is(typeof(f(x))) || is(typeof(x.f)))
Basically, is(X) checks if X has a valid type (which include void if f
doesn't return anything), and typeof(Y) returns the type of Y if it
exists, otherwise it is an error and has no type. So if f(x) doesn't
compile, then typeof(f(x)) has no type, and so is(typeof(f(x))) will be
false. Ditto for x.f.
T
--
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