C++ - overloading [] operator
The operator[]
overload will be selected based on the const
-qualification of the object you call it on.
Array<> intArray;
intArray[1]; //calls T& operator[]
const Array<> constArray;
constArray[1]; //calls T operator[]
If you remove the const
from T operator[]
, you get an error because the member functions cannot have the same const
-qualification and parameters as there would be no way to select between them.
First thing, regard []
as syntactic sugar for calling this->operator[]
.
The const
version will be called if this
is a const
pointer, else the non-const
version will be called.
Moving on, you ought to use const T& operator [](int idx) const {
, i.e. have the const
version return a const
reference. That will save the overhead of taking a deep copy.
Finally, the const
-ness of a function is part of its signature. This allows you to overload based on const
-ness. Otherwise you couldn't have the two versions of operator[]
.