On Thursday, 6 August 2020 at 16:01:35 UTC, jmh530 wrote:
The code below compiles, but I want to put an additional
constraint on the `test` function is only called with a Foo
struct.
I tried things like is(T == Foo) and is(T : Foo), but those
don't work. However, something like is(T!int : Foo!int) works,
but is(T!U == Foo!U, U) doesn't. Any idea why is(T!U == Foo!U,
U) doesn't work?
struct Foo(T)
{
T x;
}
void test(alias T)()
if (__traits(isTemplate, T))
{
import std.stdio: writeln;
writeln("there");
}
void main()
{
test!Foo();
}
`is(...)` only works on types. You're looking for
`__traits(isSame, T, Foo)`.
For `is(T!U == Foo!U, U)` to work, the compiler would have to
guess U. If the first guess doesn't work, it would have to guess
again, and again, and again, until it finds a U that does work.
Could take forever.
