does std::function reset its internal function after being moved

Question

#include <iostream>
using namespace std;

#include <functional>

        template <class F>
        class ScopeExitFunction
        {
        public:
            ScopeExitFunction(F& func) throw() :
                m_func(func)
            {
            }

            ScopeExitFunction(F&& func) throw() :
                m_func(std::move<F>(func))
            {
            }

            ScopeExitFunction(ScopeExitFunction&& other) throw() :
                m_func(std::move(other.m_func))
            {
             //  other.m_func = []{};
            }

            ~ScopeExitFunction() throw()
            {
                m_func();
            }

        private:
            F m_func;
        };

int main() {
    {
        std::function<void()> lambda = [] { cout << "called" << endl; };
        ScopeExitFunction<decltype(lambda)> f(lambda);
        ScopeExitFunction<decltype(lambda)> f2(std::move(f));
    }
    return 0;
}

without uncomment this line // other.m_func = []{}; the program produce this output :

Executing the program.... $demo called terminate called after throwing an instance of 'std::bad_function_call' what(): bad_function_call

is it normal that std::function doesn't reset his internal function when moved ?

Answer 1

According to C++11 20.8.11.2.1/6 [func.wrap.func.con], move-constructing from an existing std::function object leaves the original object "in a valid state with an unspecified value". So basically don't assume anything. Just don't use the original function object, or don't move from it if you still need it.

Answer 2

is it normal that std::function doesn't reset his internal function when moved ?

On the contrary, in your case it does reset the internal function, meaning the internal handle is set to zero, and the std::function is not a real function any more. And since that happened the call to operator() does not go well.