std::bind with std::reference_wrapper<T> argument doesn't work on function object with template operator() by Sven Groot

In order to pass arguments by reference to the std::bind function, you use std::reference_wrapper<T>, typically by invoking std::ref.

However, if std::bind is used on a function object whose operator() is itself a template function, that template will be instantiated with std::reference_wrapper<T> rather than T itself, which is incorrect behavior.

For example, the following code does not compile:

#include <iostream>
#include <functional>
using namespace std;

struct Foo
{
typedef void result_type;

template<typename T>
void operator()(T &n)
{
    n = 5;
}
};

int main()
{
int n = 0;
Foo f;
auto func = bind(f, ref(n));

func();
cout << n << endl;

return 0;
}

This code does compile and work correctly on g++ 4.6.1, or if std::bind/std::ref is replaced with boost::bind/boost::ref.

A real-world example where this is a problem is the Boost.Random library. For example, the following does not work on VC++ (but does on g++):

boost::mt19337 engine;
boost::uniform_int<> distribution(0, 10);
auto func = std::bind(distribution, std::ref(engine));