C++11 does not deduce type when std::function or lambda functions are involved

Question

When I define this function,

template
set test(const set& input) {
    return input;
}

I can call it u

Answer 1

If we have:

template 
int myfunc(std::function lambda)
{
  return lambda(2);
}

int r = myfunc([](int i) { return i + 1; });

It will not compile. But if you previously declare:

template 
static auto getFuncType(Func* func = nullptr, Arg1* arg1 = nullptr) -> decltype((*func)(*arg1));

template 
int myfunc(Func lambda)
{
  return myfunc())>(lambda);
}

You can call your function with a lambda parameter without problem.

There are 2 new pieces of code here.

First, we have a function declaration which is only useful to return an old-style function pointer type, based on given template parameters:

template 
static auto getFuncType(Func* func = nullptr, Arg1* arg1 = nullptr) -> decltype((*func)(*arg1)) {};

Second, we have a function that takes a template argument to build our expected lambda type calling 'getFuncType':

template 
int myfunc(Func lambda)
{
  return myfunc())>(lambda);
}

With the correct template parameters, now we can call the real 'myfunc'. Complete code will be:

template 
int myfunc(std::function lambda)
{
  return lambda(2);
}

template 
static auto getFuncType(Func* func = nullptr, Arg1* arg1 = nullptr) -> decltype((*func)(*arg1)) {};

template 
int myfunc(Func lambda)
{
  return myfunc())>(lambda);
}

int r = myfunc([](int i) { return i + 1; });

You can declare any overload for 'getFuncType' to match your lambda parameter. For example:

template 
static auto getFuncType(Func* func = nullptr, Arg1* arg1 = nullptr, Arg2* arg2 = nullptr) -> decltype((*func)(*arg1, *arg2)) {};