overload resolution of function templates

Question

Question is the code. It looks like the 2nd function is more special than the 1st one. Why the more general one is called in the following code? How can I make the other function to be used?

template <typename T>
class Base{
public:
    Base(){}
    void print() const {cout<<"Base class"<<endl;}
};

template <typename T>
class Derived :public Base<T>{
public:
    Derived() {}
    void print() const {cout<<"Derived class"<<endl;}
};

template <typename T>
void func(T x){    // <----- Why is function is called?
    x.print();
    cout<<"in func(T)"<<endl;
}

template <typename T>
void func(const Base<T>& x){
    x.print();
    cout<<"in func(Base<T>)"<<endl;
}

int main () {
    Base<int> b;
    Derived<int> d;
    func(d);
    return 0;
}

Note that I am passing the Derived object to the function.

This post can help you: http://stackoverflow.com/questions/22411482/c-template-functions-overload-resolution. See the answer of @NikosAthanasiou. — chema989, Jun 22 '16 at 02:32
Possible duplicate of: http://stackoverflow.com/questions/31563580/c-templated-function-overloading-rules?lq=1 — sameerkn, Jun 22 '16 at 06:51

score 0 · Answer 1 · edited Jun 22 '16 at 05:36

Note that I am passing the Derived object to the function.

In this case, there must be one implicit conversion (from Derived<T> to Base<T>) for the argument of the 2nd function template. While for the 1st function template it's an exact match and preferred.

How can I make the other function to be used?

You can change the parameter type of the 2nd function template to avoid implicit conversion. You can also use std::enable_if with std::is_base_of to make it only work with the base class and its derived classes.

template <typename T, template <typename> class D>
typename std::enable_if<std::is_base_of<Base<T>, D<T>>::value>::type
func(const D<T>& x){
    x.print();
    cout<<"in func(Base<T>)"<<endl;
}

Live Demo

^{BTW: I think Base::print() should be a virtual function.}

sameerkn · Answer 2 · 2016-06-22T06:45:12.770

0

Simple answer would be:

template instantiation are done at compile time: At compile time there is no function which can take Derived<int>, therefore void func(T x) is used to instantiate the definition of that kind of function.

implicit dynamic conversion are done at run time.

edited Jun 22 '16 at 06:45

answered Jun 22 '16 at 06:37

sameerkn

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overload resolution of function templates

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