index operator constness

Question

Why do we need two? Under which circumstance each of the following operator[]s are called?

  class X {
      public:
        //...
        int &operator [](int index);
        const int &operator [](int index) const;
    };

Answer 1

If the class X instance is const the const version is called, otherwise the non-const:

void function1( X& x )
{
    x[0]; // x is non-const, non-const version is called
    x[0] = ... // allright
}

void function2( const X& x )
{
    x[0]; // x is const, const version is called
    x[0] = ... // illegal - won't compile
}

this is done to facilitate const-correctness.

Answer 2

The const one would be used if operator[] were invoked on a const X object; otherwise the non-const one.

X x;
x[42] = 42; // ok

X x_const;
x_const[42] = 42; // error: cannot assign to const reference

Answer 3

The first one is called when the instance of the class is referred to as a non-const type. The second is called when the type that refers to the instance is const.

For example:

void check_array(X const& x)
{
   // x[0] = 2;  // That would be a compile error.  
                 // And it's GOOD that we have such a check!
   return x[1] == 5; //calls non-const version
}

void mutate_array(X& x)
{
   x[0] = 2; // calls non-const version
   if (x[1] == 5){  //calls non-const version anyway!
       X const& const_ref = x;
       return const_ref[1] == 5;  //NOW it calls const version.
   }
}

The reasons to use it are:

• to force a check that we don't try to modify something, that shouldn't be modified (as in check_array() function)
• to allow compiler to optimize code, knowing that certain values won't change in a block.

Answer 4

Using 2 versions offers additional ways for the compiler to optimize. No value can't be assigned to const int &. The second operator is called whenever the statements allow it and since the operator is const, the compiler has additional possiblities for optimization. Whenever a result is assigned to the return value of that operator, the compiler selects the first version (which is not const).

For a more complete explanation - see Scott Meyers (More) Effective C++

Answer 5

foo( X x )
{
  x[0];  // non-const indexer is called
}

bar ( const X x )
{
  x[0]; //const indexer is called
}