Question

In a project I maintain, I see a lot of code like this for simple get/set methods

const int & MyClass::getFoo() { return m_foo; }  void MyClass::setFoo(const int & foo) { m_foo = foo; } 

What is the point in doing that instead of the following?

int MyClass::getFoo() { return m_foo; }  // Removed 'const' and '&'   void MyClass::setFoo(const int foo) { m_foo = foo; }  // Removed '&' 

Passing a reference to a primitive type should require the same (or more) effort as passing the type’s value itself, right?
It’s just a number after all…
Is this just some attempted micro-optimization or is there a true benefit?

Answer 1

The difference is that if you get that result into a reference yourself you can track the changes of the integer member variable in your own variable name without recalling the function.

const &int x = myObject.getFoo(); cout<<x<<endl; //... cout<<x<<endl;//x might have changed 

It’s probably not the best design choice, and it’s very dangerous to return a reference (const or not), in case a variable that gets freed from scope is returned. So if you return a reference, be careful to be sure it is not a variable that goes out of scope.

There is a slight difference for the modifier too, but again probably not something that is worth doing or that was intended.

void test1(int x) {   cout<<x<<endl;//prints 1 }  void test2(const int &x) {   cout<<x<<endl;//prints 1 or something else possibly, another thread could have changed x }  int main(int argc, char**argv) {   int x = 1;   test1(x);   //...   test2(x);   return 0; } 

So the end result is that you obtain changes even after the parameters are passed.