The following code fails to print anything, although I believe it should be valid and should print `const(int)`. shared const int i; static if (is(typeof(i) == shared U, U)) { pragma(msg, U); } void main() { }
No, because you use ==, which does exact matching. If you want to partially match the type (i.e. check if it's implicitly convertible to shared), use `:`, not `==`: shared const int i; static if (is(typeof(i) == shared U, U)) { pragma(msg, U); // now prints const(int) } void main() { }
I'm assuming in your example you meant to use : instead of ==. That aside, why shouldn't this work? I'm trying to destructure some type which matches the pattern `shared U`. `shared const(int)` *should* match that pattern. What I'm trying to express is `∃ U: shared U == shared const int`; it seems incorrect to me that that I should have to use the subtyping form of is.
(In reply to monkeyworks12 from comment #2) > I'm assuming in your example you meant to use : instead of ==. Oops, yes. > That aside, why shouldn't this work? I'm trying to destructure some type > which matches the pattern `shared U`. `shared const(int)` *should* match > that pattern. What I'm trying to express is `∃ U: shared U == shared const > int`; it seems incorrect to me that that I should have to use the subtyping > form of is. OK, I see what you mean now. It looks like partial destructuring of type qualifiers was never implemented for == variants of IsExpression. It is the same for inout: static assert(!is(shared const int == shared U, U)); // should work static assert( is(shared const int : shared U, U)); static assert(!is(inout const int == const U, U)); // should work static assert( is(inout const int : const U, U));
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