A const std::function wraps a non-const operator() / mutable lambda

Question

Consider the following example:

#include 
#include 

struct A
{
    int i;
    void operator()() 
    {
        std::cout &         


        

Answer 1

Is it normal that a const std::function may wrap a mutable functor?

Unfortunately, yes. std::function::operator() is unconditionally qualified as const and doesn't care whether or not the wrapped Callable is mutated. Some papers attempted to tackle this issue, but AFAIK nothing concrete was yet decided:

• http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/n4348.html

• http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2017/p0045r1.pdf