CODEWITHSUNDEEP
c++
Sukhanov Niсkolay
Sukhanov Niсkolay

Reputation: 1328

Function template specialization without templated argument

Let's look on a template function which take void argument and returns void:

template<class T>
void f() {
    cout << "template" << endl;
}

Later we specialize this function:

template<>
void f<int> () {
    cout << "int" << endl;
}
template<>
void f<double> () {
    cout << "double" << endl;
}

int main() {
    f<int> ();
    f<double> ();
}

The question is why this compiles? We have three function with the same signature: void(void), and I expected that it should broke with multiple declaration?

Upvotes: 2

Views: 272

Answers (2)

TartanLlama
TartanLlama

Reputation: 65770

This is because you are explicitly stating which function to use. f<int>() can not map to f<double>() because of the difference in template argument; i.e. this is a completely unambiguous call.

Upvotes: 6

John Zwinck
John Zwinck

Reputation: 249642

Why shouldn't it work? Those are three different implementations, depending on the template parameter. Each different value you use when instantiating a template creates a completely new copy of the entire templated thing. So here you make three functions, they're all unique and it works fine.

By the way, the symbol name won't be just f, look up C++ name mangling.

Upvotes: 2

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