std::function, template parameter T(X)

Question

I'm wondering if anyone can please explain how, given types T and X, std::function takes T(X) as a template parameter.

int(double) looks like the usual cast from double to int, so how is std::function parsing it as distinct types? I did search but didn't find anything that specifically addresses this question. Thanks!

Answer 1

It can use partial template specialization. Look at this:

template <typename T>
class Func;

template <typename R, typename... Args>
class Func<R(Args...)>
{
public:
    Func(R(*fptr)(Args...)) {/*do something with fptr*/}
};

This class takes a single template parameter. But unless it matches R(Args...) (i.e. a function type returning R and taking zero or more args), there wont be a definition for the class.

int main() { Func<int> f; }
// error: aggregate 'Func<int> f' has incomplete type and cannot be defined

.

int func(double a) { return a+2; }
int main() { Func<int(double)> f = func; }
// ok

The specialization can now operate on R and Args to do its magic.

Note that int(double) is a function type. Since you cannot create raw function objects, you don't usually see this syntax outside the template world. If T is int(double) then T* is a function pointer just like int(*)(double).