Deduce lambda expression in a template function call C++ [duplicate] - c++

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.

Answer 2

As has been revealed by other posters, this is a template argument deduction for std::function.
One intuitive way to make the second code snippet work is to add your base type when calling the template function: findMatches<int>.
Another way not mentioned by Xeo is using std::is_convertible:
template<typename BaseT, typename FUNC>
vector<BaseT> findMatches(vector<BaseT> search, function <bool (const BaseT &)> func)
{
static_assert(std::is_convertible<FUNC, function<bool (const BaseT &)> >::value, "func must be convertible to ...");
vector<BaseT> tmp;
for(auto item : search)
{
if( func(item) )
{
tmp.push_back(item);
}
}
return tmp;
}
It avoids wrapping lamda into std::function, and provides a cleaner error message.