Why can't I pass a lambda to this function which takes a std::function? [duplicate]

Question

The following program is illegal, and I would like to understand why:

#include <functional>
#include <iostream>

template<typename Result, typename Arg>
void deduce(std::function<Result(Arg)> f)
{
  std::cout << "Result: " << typeid(Result).name() << std::endl;
  std::cout << "Arg: " << typeid(Arg).name() << std::endl;
}


int main()
{
  auto f = [](int x)
  {
    return x + 1;
  };

  deduce(f);

  return 0;
}

clang's output:

$ clang -std=c++11 test.cpp 
test.cpp:48:3: error: no matching function for call to 'deduce'
  deduce(f);
  ^~~~~~
test.cpp:26:6: note: candidate template ignored: could not match 'function<type-parameter-0-1 (type-parameter-0-0)>' against '<lambda at test.cpp:34:13>'
void deduce(std::function<T2(T1)> f)
     ^
1 error generated.

It seems like I ought to be able to convert my lambda to the std::function received by deduce. Why is it not possible for the compiler to apply an appropriate conversion in this case?

Answer 1

The problem is that while a lambda that takes an int and returns an int can be converted into a std::function<int(int)>, its type is not std::function<int(int)> but an arbitrary implementation-defined type I think.

You can work around this by telling the compiler what types you want. Then the conversion will happen as expected.

auto f = [](int x){ return x + 1; };
deduce<int, int>(f);  // now ok

Alternatively, be explicit on the static type of f.

std::function<int(int)> f = [](int x){ return x + 1; };
deduce(f);  // now also ok