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

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?

CodePudding user response：

It behaves like

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

If you want to get errors, declare them like

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