What does 'const' do in operator() overloading?

23092019  
Question
I have a code base, in which for Matrix class, these two definitions are there for ()
operator:
template <class T> T& Matrix<T>::operator() (unsigned row, unsigned col)
{
......
}
template <class T> T Matrix<T>::operator() (unsigned row, unsigned col) const
{
......
}
One thing I understand is that the second one does not return the reference but what does const
mean in the second declaration? Also which function is called when I do say mat(i,j)
?
Solution
Which function is called depends on whether the instance is const or not. The first version allows you to modify the instance:
Matrix<int> matrix;
matrix(0, 0) = 10;
The const overload allows readonly access if you have a const instance (reference) of Matrix:
void foo(const Matrix<int>& m)
{
int i = m(0, 0);
//...
//m(1, 2) = 4; //won't compile
}
The second one doesn't return a reference since the intention is to disallow modifying the object (you get a copy of the value and therefore can't modify the matrix instance).
Here T is supposed to be a simple numeric type which is cheap(er) to return by value. If T might also be a more complex userdefined type, it would also be common for const overloads to return a const reference:
template <class T>
class MyContainer
{
//..,
T& operator[](size_t);
const T& operator[](size_t) const;
}
OTHER TIPS
The const version will be called on const Matrices. On nonconst matrices the nonconst version will be called.
Matrix<int> M;
int i = M(1,2); // Calls nonconst version since M is not const
M(1,2) = 7; // Calls nonconst version since M is not const
const Matrix<int> MConst;
int j = MConst(1,2); // Calls const version since MConst is const
MConst(1,2) = 4; // Calls the const version since MConst is const.
// Probably shouldn't compile .. but might since return value is
// T not const T.
int get_first( const Matrix<int> & m )
{
return m(0,0); // Calls the const version as m is const reference
}
int set_first( Matrix<int> & m )
{
m(0,0) = 1; // Calls the nonconst version as m is not const
}
Which function is called depends on whether the object is const
. For const
objects const
overload is called:
const Matrix<...> mat;
const Matrix<...>& matRef = mat;
mat( i, j);//const overload is called;
matRef(i, j); //const overloadis called
Matrix<...> mat2;
mat2(i,j);//nonconst is called
Matrix<...>& mat2Ref = mat2;
mat2Ref(i,j);//nonconst is called
const Matrix<...>& mat2ConstRef = mat2;
mat2ConstRef(i,j);// const is called
The same applies to pointers. If the call is done via a pointertoconst, a const overload is called. Otherwise a nonconst overload is called.