How to avoid code duplication in static if branches?

[Andrej Mitrovic]

You're right it should be able do that dead-code elimination thing.
Slipped my mind. :)

On 3/4/12, Daniel Murphy <yebblies at nospamgmail.com> wrote:
> "Andrej Mitrovic" <andrej.mitrovich at gmail.com> wrote in message
> news:mailman.364.1330825349.24984.digitalmars-d-learn at 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.
>
>
>