C++ selectively overloading [ ]? - c++

Question

Given:
class example
{
public:
std::vector<std::vector<int>> a;
int b;
}
func()
{
example e;
... // populate e
I could then use examples members like so.
int n = e.b;
int n2 = e.a[2][3];
However, could I alternatively override the [ ] operator such that.
int n = e.b;
int n2 = e[2][3];
}
?
edit: Sorry, the example is now fixed.

Answer 1

What you could do is overload the access operator and delegate it to the vector:
class example
{
public:
std::vector<std::vector<int>> a;
int b;
const std::vector<int>& operator[](std::size_t i) const
{
return a[i];
}
std::vector<int>& operator[](std::size_t i)
{
return a[i];
}
};
The first [] will then return a reference to a's respective element, on which the second [] will be used.

Answer 2

One naive solution would be to define your own operator[]:
class example
{
std::vector<std::vector<int>> v;
public:
std::vector<int> const & operator[](std::size_t i) const { return v[i]; }
std::vector<int> & operator[](std::size_t i) { return v[i]; }
// ...
};
Now if you have example e;, then e[1] is a vector of ints, etc.
It looks like you want to write some matrix class, though. For that, it's more efficient to have just a single vector and access it in strides:
class Matrix
{
std::size_t cols;
std::size_t rows;
std::vector<int> v;
public:
explicit Matrix(std::size_t r, std::size_t c) : cols(c), rows(r), v(r*c) { }
int operator()(std::size_t i, std::size_t j) const { return v[i * cols + j]; }
int & operator()(std::size_t i, std::size_t j) { return v[i * cols + j]; }
};
Now you can say: Matrix m(4, 7); m(1, 3) = 8;. You have to use the round-bracket operator for that, since the square-bracket operator must always take exactly one argument and thus isn't suitable here.