You're right it should be able do that dead-code elimination thing. Slipped my mind. :)
On 3/4/12, Daniel Murphy <yebbl...@nospamgmail.com> wrote: > "Andrej Mitrovic" <andrej.mitrov...@gmail.com> wrote in message > news:mailman.364.1330825349.24984.digitalmars-d-le...@puremagic.com... >> import std.stdio; >> void check() { writeln("check"); } >> >> struct Foo { bool isTrue = true; } >> struct Bar { } >> >> void test(T)(T t) >> { >> static if (is(T == Foo)) >> { >> if (t.isTrue) >> check(); >> } >> else >> { >> check(); >> } >> } >> >> void main() >> { >> Foo foo; >> Bar bar; >> test(foo); >> test(bar); >> } >> >> I want to avoid writing "check()" twice. I only have to statically >> check a field of a member if it's of a certain type (Foo). >> >> One solution would be to use a boolean: >> void test(T)(T t) >> { >> bool isTrue = true; >> static if (is(T == Foo)) >> isTrue = t.isTrue; >> >> if (isTrue) >> check(); >> } >> >> But that kind of defeats the purpose of static if (avoiding runtime >> overhead). Does anyone have a trick up their sleeve for these types of >> situations? :) > > Have you checked the generated code? When the static if check fails, it > should be reduced to: >> void test(T)(T t) >> { >> bool isTrue = true; >> >> if (isTrue) >> check(); >> } > > And the compiler should be able to tell that isTrue is always true. > > Otherwise, > > void test(T)(T t) > { > enum doCheck = is(T == Foo); > bool isTrue = true; > static if (is(T == Foo)) > auto isTrue = t.isTrue; > if (!doCheck || isTrue) > check(); > } > > The compiler takes care of it because doCheck is known at compile time. > > >
