C++11 lambda can be assigned to std::function with incorrect signature

Question

The following compiles and runs (under Apple LLVM version 6.1.0 and Visual C++ 2015):

#include <functional>
#include <iostream>

struct s { int x; };

int main(int argc, char **argv)
{
    std::function<void (s &&)> f = [](const s &p) { std::cout << p.x; };
    f(s {1});
    return 0;
}

Why doesn't the assignment std::function<void (s &&)> f = [](const s &p) { std::cout << px; }; std::function<void (s &&)> f = [](const s &p) { std::cout << px; }; generate an error? A function accepting an rvalue reference should not have the same signature as a function accepting a const lvalue reference, should it? Dropping the const from the lambda's declaration does generate an error as expected.

Answer 1

To expand on the existing comment and answer:

The point of std::function<R(A...)> is that it can wrap any function or functor that can be called with A... and have the result stored in an R .

So, for example,

std::function<int(int)> f = [](long l) { return l; };

is just peachy.

So what you have to ask yourself when you see something like this: if you have a lambda taking const T & , and you have an expression of type T && (or, more accurately, you have an xvalue of type T ), can you use that expression to call the lambda?

Yes, you can.

And if you can, then std::function is supposed to be able to store that functor. That's pretty much the main point of std::function .

Answer 2

Please take this with a grain of salt. This is what I understand, but I am not sure.

Consider the following output:

int main(int argc, char **argv)
{
  std::cout <<  std::is_convertible<s &&, s&>::value << std::endl;       //false                                                      
  std::cout <<  std::is_convertible<s &&, const s&>::value << std::endl; //true                                                     

  std::cout <<  std::is_convertible<const s &, s&&>::value << std::endl; //false                                                     
  return 0;
}

This shows that it is possible to convert a s && to a const s& . This is why the std::function 's assignment is ok.

Dropping the const from the lambda's declaration does generate an error as expected.

Indeed this is because (as shown before), converting a s && to a s & is not possible.

In the same way, trying the opposite:

std::function<void (const s &)> f = [](s &&p) { std::cout << px; }; would fail because it is not possible to convert a const s& to a s && .