Why and how is std::function<void()> allowed to be assigned to callables with return types in C++?

The following code is legal C++:

int m() { return 42; }

int main() {
   std::function<void()> func, func2;
   
   func = m;
   func2 = [&]() -> std::string { return "This does not return void"; };
}

By passing void() to std::function‘s template argument list, I assumed that meant func and func2 must be assigned to functions/lambdas that return nothing. That is clearly not true.

Firstly, why is std::function designed to do this? This seems really surprising to me.

Secondly, how did they design std::function this way? I know std::function uses type erasure to be assignable to different types of callables, but since we explicitly supplied void(), shouldn’t the different callables assigned to func and func2– whether they are lambdas or functions – only be returning void?

>Solution :

It behaves like

class function {
 public:
  void operator() {
    m();
  }
};

If you want to get errors, declare them like

std::function<void*()> func, func2;