hasUDA with this
jmh530
john.michael.hall at gmail.com
Fri Jul 20 14:11:23 UTC 2018
On Friday, 20 July 2018 at 07:25:23 UTC, Simen Kjærås wrote:
>
> UDAs apply to symbols. When you pass something to a function,
> the symbol does not go with the value, and thus that UDA goes
> away. It's similar to assigning to a different variable:
>
> import std.traits : hasUDA;
> @("foo")
> Foo foo1;
> Foo foo2 = foo1;
> assert(!hasUDA!(foo2, "foo"));
>
> Since UDAs are compile-time constructs, they really can't
> follow values around in that way. Consider:
>
> void main(string[] args) {
> @("foo")
> Foo foo1;
> @("bar")
> Foo foo2;
> Foo foo3 = (args.length % 2) ? foo1 : foo2;
> assert(hasUDA!(foo3, "foo")); // Would you expect this to
> pass or fail?
> }
>
> Timoses first checkUDA function is exactly what I would suggest
> you use.
>
> --
> Simen
That the UDAs don't copy with the values potentially throws my
idea into the crapper...
Anyway, what I really wanted to do is change behavior of the
struct, at compile-time, based on a UDA. For instance, what I
really want is more like:
struct Foo
{
static if (hasUDA!(this, DisableCasting))
{
@disable T opCast(T) { }
}
}
So that if the instance has the particular UDA, then some kind of
error would be thrown at compile-time that opCast was disabled. I
don't really need anything at run-time. Obviously, I could do
this as a template with a compile-time switch. I just wanted to
see if it was possible with a UDA because then I could apply it
to many different structs by just adding a template mixin of the
above. I wouldn't need to make them different types. All
functions that currently use them would work, unless there is a
cast, in which case you know that at compile-time.
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