"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.
