如何在以下代码中解决函数重载歧义

Question

Suppose there are function ovelaods for int and double .

void Print(int v)
{
    cout << v << endl;
}

void Print(double v)
{
    cout << v << endl;
}

There is function wich is intended to pass in as callable one of above functions.

template<typename FnT>
void Function(int v, FnT&& fn)
{
    fn(v);
}

But following code is ambiguous:

Function(1, Print);

Live example

Compiller fails to deduce type for second argument. This can be easily solved:

Function(1, static_cast<void(*)(int)>(Print));

I believe exists more generic and elegant way to solve the issue. The issue can be raised up in STL algorithm, when function overload is added.

vector<int> v = { 1,2,3 };
for_each(v.begin(), v.end(), Print); // ambigous code

How to solve this in pretty way?

Answer 1

Wrap it in a function object:

#include <iostream>
#include <utility>

void Print(int v)
{
    std::cout << v << std::endl;
}

void Print(double v)
{
    std::cout << v << std::endl;
}

template<typename FnT>
void Function(int v, FnT&& fn)
{
    fn(v);
}

auto print_wrapper() {
    return [](auto&&...args) -> decltype(auto)
    {
        return Print(std::forward<decltype(args)>(args)...);
    };
}

int main()
{
    Function(1, print_wrapper());
}

On a more theoretical note, what we're doing here is using a bandage to fix a broken design.

A better design would be to define the concept of Print as a template function object. Then we can specialise it and customise it to our heart's content.

A (very complete) model of this is boost::hash<> which is worth half a day of anyone's time as a study exercise.

A simple example:

#include <iostream>
#include <utility>
#include <string>

struct Printer
{
    void operator()(const int& i) const {
        std::cout << i << std::endl;
    }

    void operator()(const double& i) const {
        std::cout << i << std::endl;
    }

    template<class Anything>
    void operator()(const Anything& i) const {
        custom_print(i);
    }
};

struct Foo
{

};
void custom_print(Foo const& f) 
{
    std::cout << "a Foo" << std::endl;
}

template<typename X, typename FnT>
void Function(X const& x, FnT&& fn)
{
    fn(x);
}

int main()
{
    Function(1, Printer());
    Function(1.0, Printer());
    Function(Foo(), Printer());
}

Answer 2

另外，您可以使用lambda：

Function(42, [](int i) {Print(i);});.