c++ - Lambdas and std::function

Question

I'm trying to catch up on C++11 and all the great new features. I'm a bit stuck on lambdas.

Here's the code I was able to get to work:

#include <iostream>
#include <cstdlib>
#include <vector>
#include <string>
#include <functional>

using namespace std;

template<typename BaseT, typename Func>
vector<BaseT> findMatches(vector<BaseT> search, Func func)
{
    vector<BaseT> tmp;

    for(auto item : search)
    {
        if( func(item) )
        {
            tmp.push_back(item);
        }
    }

    return tmp;
}

void Lambdas()
{
    vector<int> testv = { 1, 2, 3, 4, 5, 6, 7 };

    auto result = findMatches(testv, [] (const int &x) { return x % 2 == 0; });

    for(auto i : result)
    {
        cout << i << endl;
    }
}

int main(int argc, char* argv[])
{

    Lambdas();

    return EXIT_SUCCESS;
}

What I would like to have is this:

template<typename BaseT>
vector<BaseT> findMatches(vector<BaseT> search, function <bool (const BaseT &)> func)
{
    vector<BaseT> tmp;

    for(auto item : search)
    {
        if( func(item) )
        {
            tmp.push_back(item);
        }
    }

    return tmp;
}

Basically I want to narrow down the possible lambdas to a sensible subset of functions. What am I missing? Is this even possible? I'm using GCC/G++ 4.6.

Answer 1

Stephan T. Lavavej explains why this doesn't work in this video. Basically, the problem is that the compiler tries to deduce BaseT from both the std::vector and the std::function parameter. A lambda in C++ is not of type std::function, it's an unnamed, unique non-union type that is convertible to a function pointer if it doesn't have a capture list (empty []). On the other hand, a std::function object can be created from any possible type of callable entity (function pointers, member function pointers, function objects).

Note that I personally don't understand why you would want to limit the incoming functors to that specific signature (in addition to the fact that indirection through a polymorphic function wrapper, like std::function, is by far more inefficient than a direct call to a functor (which may even be inlined)), but here's a working version. Basically, it disables argument deduction on the std::function part, and only deduces BaseT from the std::vector argument:

template<class T>
struct Identity{
  typedef T type;
};

template<typename BaseT>
vector<BaseT> findMatches(vector<BaseT> search, 
    typename Identity<function<bool (const BaseT &)>>::type func)
{
    vector<BaseT> tmp;

    for(auto item : search)
    {
        if( func(item) )
        {
            tmp.push_back(item);
        }
    }

    return tmp;
}

Live example on Ideone.

Another possible way would be to not restrict the functor type directly, but indirectly through SFINAE:

template<class T, class F>
auto f(std::vector<T> v, F fun)
    -> decltype(bool(fun(v[0])), void())
{
  // ...
}

Live example on Ideone.

This function will be removed from the overload set if fun doesn't take an argument of type T& or if the return type is not convertible to bool. The , void() makes f's return type void.