On Monday, 4 January 2021 at 14:40:31 UTC, ag0aep6g wrote:
On 04.01.21 15:37, Ola Fosheim Grøstad wrote:
On Monday, 4 January 2021 at 14:11:28 UTC, ag0aep6g wrote:
`Bar!int` is an alias. It's indistinguishable from `Foo!int`. The code fails in the same manner when you replace "Bar!int" with "Foo!int".


Wrong. This succeeds:

  struct Foo(T) {}

   alias Bar(T) = Foo!T;

   void f(T)(Foo!T x) {}

   void main() {
     f(Bar!int());
   }

You didn't replace "Bar!int" with "Foo!int". You replaced "Bar!T" with "Foo!T". That's something else entirely.

IMO, this is a better example, even if it's a little more verbose.

struct Foo(T) {}

alias Bar(T) = Foo!T;

void f(T)(Bar!T x) {}

void main() {
    auto x = Bar!int();
    f(x);
}

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