Question

I wanted to create these function templates you see below. Their purpose is to compare functors, but I needed to cover a special case for the boost.bind type of functors.

template<typename R, typename F, typename L>
void compare(boost::_bi::bind_t<R, F, L>& lhs, boost::_bi::bind_t<R, F, L>& rhs)
{
    std::cout << lhs.compare(rhs) << std::endl;
}

template<typename T>
void compare(T lhs, T rhs)
{
    std::cout << (lhs == rhs) << std::endl;
}

The problem is that when I do compare(boost::bind(func, 1), boost::bind(func, 1)), the compiler tries to use the second template. If I comment out the second one, it will correctly use the one specialized for the boost.bind type and everything will work fine.

How can I make it choose the correct function template to use?

Answer 1

boost::bind returns a value which can’t be bound to a non-const reference. Your better specialized template needs to take it’s arguments by value or by const reference otherwise it won’t be considered in the call: compare( boost::bind(func, 1), boost::bind(func, 1) ).

This test program compiles and works correctly on my platform.

#include <boost/bind/bind.hpp>
#include <iostream>
#include <ostream>

template<typename R, typename F, typename L>
void compare(const boost::_bi::bind_t<R, F, L>& lhs
                 , const boost::_bi::bind_t<R, F, L>& rhs)
{
    std::cout << lhs.compare(rhs) << std::endl;
}

template<typename T>
void compare(T lhs, T rhs)
{
    std::cout << (lhs == rhs) << std::endl;
}

void func(int) {}

int main()
{
    compare( boost::bind(func, 1), boost::bind(func, 1) );
}