Assigning std::function<int(int)> to std::function<const int&(const int& x)>

Question

The following code compiles but yields undefined output in VC++ 2015 (release) and a runtime error with other compilers.

#include <functional>
#include <iostream>
int main()
{
    std::function<int(int)> f = [](int x) { return x; };
    std::function<const int&(const int& x)> g = f;
    std::cout << g( 42 ) << std::endl;
}

Why is the assignment g = f; allowed?

Answer 1

Consider the equivalent code rewritten to avoid lambdas or std::function:

int f(int x) { return x; } 
int const& g(int const& x) { return f(x); } 

This is perfectly well-formed code, that nevertheless returns a dangling reference to a temporary and thus will end up causing undefined behavior. The original code is vaid for the same reason: you can implicitly convert an object to a reference of the same type. Unfortunate, in this case.

Answer 2

An rvalue can be bound to a const&. A const& can be converted to an rvalue.

Examine this:

int f(int x){return x;}
int const& g(int const& x){ return f(x); }

Similarly, the call to g is legal, there are no errors, yet reading the result of g(42) is UB -- the reference dangles.

A good compiler will see the reference bound to temporary being returned and warn.

function simply checks if the types can be converted between; it does no lifetime analysis. Possibly it should, as we can detect this error statically.