enable_if and conversion operator?

Question

Any chance to use enable_if with a type conversion operator? Seems tricky, since both return type and parameters list are implicit.

Answer 1

From the little research I did (and ignoring the c++0x comment from Johannes), my answer is that it depends what you want the enable_if for. If you want the conversion operation to T to exist or not from the type T then it seems that the answer is no, there is no way in C++03 (as Ugo said). But if you need the enable_if to change the behavior of the operator depending on the type of T then yes, there is a workaround which is to call an enabled helper function (called to as Matthieu suggested).

#include
#include
#include

struct B{
    B(const B& other){}
    B(){}
};

struct A{
    template
    T to(typename boost::enable_if_c::value, void*>::type = 0){
        std::clog << "converted to non class" << std::endl;
        return T(0);
    }
    template
    T to(typename boost::enable_if_c::value, void*>::type = 0){
        std::clog << "conveted to class" << std::endl;
        return T();
    }
    template
    operator T(){
        return to();
    }
};

int main(){
    A a;
    double d = (double)a; // output: "converted to non class"
    B b = (B)(a); // output: "converted to class"
    return 0;
}

For the record, I was frustrated with this for several days, until I realized that I wanted enable_if not for SFINAE but for compile-time behavior change. You may also find that this is the real reason for your need for enable_if also. Just a suggestion.

(Please note that this is an answer for the C++98 era)