按值返回常数和非常数

Question

There is such code:

   const int fun(){ return 2; } // can be assigned to int and const int
   int fun2(){ return 2; } // can be assigned to int and const int

Is there any difference in using these functions? They both return by value so it is always copied at the end of function call.

Answer 1

Is there any difference in using these functions?

No. There is, however, a difference in their type , and if the functions returned a class type, there would be difference regarding invoking methods on the return value .

Answer 2

There's no practical difference when returning an int , basically because anything you do with a temporary of builtin type only needs its value. You can take a const reference to a temporary - it may be valid (if unwise) in the latter case to cast that const reference to non-const and modify the temporary through it, but I can't be bothered to look up whether temporaries of builtin type really are mutable, and there's not any great practical need to do anything like that.

When returning a class type there is a difference - in the second case you can call a non-const member function on the function's return value, and in the first case you can't. For example, given std::string fun2() { return "hello"; } std::string fun2() { return "hello"; } you can do std::cout << (fun2() += " world\\n"); , or std::string s("foo"); std::cout << s; fun2().swap(s); std::cout << "s"; std::string s("foo"); std::cout << s; fun2().swap(s); std::cout << "s"; . Such tricks are potential optimizations (especially before C++11 move semantics came along), and they don't work if fun2 returns const std::string . The second trick is called "swaptimization", which at least tells you that it's used enough to be worth naming.

Answer 3

There is no difference in using these functions. Note that your assumption that copying would happen may not be true in the face of an optimizer, that would probably inline the value 2 at the call site.