Operator[][] overload

Question

Is it possible to overload [] operator twice? To allow, something like this: function[3][3](like in a two dimensional array).

If it is pos

Answer 1

Using C++11 and the Standard Library you can make a very nice two-dimensional array in a single line of code:

std::array<std::array<int, columnCount>, rowCount> myMatrix {0};

std::array<std::array<std::string, columnCount>, rowCount> myStringMatrix;

std::array<std::array<Widget, columnCount>, rowCount> myWidgetMatrix;

By deciding the inner matrix represents rows, you access the matrix with an myMatrix[y][x] syntax:

myMatrix[0][0] = 1;
myMatrix[0][3] = 2;
myMatrix[3][4] = 3;

std::cout << myMatrix[3][4]; // outputs 3

myStringMatrix[2][4] = "foo";
myWidgetMatrix[1][5].doTheStuff();

And you can use ranged-for for output:

for (const auto &row : myMatrix) {
  for (const auto &elem : row) {
    std::cout << elem << " ";
  }
  std::cout << std::endl;
}

(Deciding the inner array represents columns would allow for an foo[x][y] syntax but you'd need to use clumsier for(;;) loops to display output.)

Answer 2

One approach is using std::pair<int,int>:

class Array2D
{
    int** m_p2dArray;
public:
    int operator[](const std::pair<int,int>& Index)
    {
       return m_p2dArray[Index.first][Index.second];
    }
};

int main()
{
    Array2D theArray;
    pair<int, int> theIndex(2,3);
    int nValue;
    nValue = theArray[theIndex];
}

Of course, you may typedef the pair<int,int>

Answer 3

Sample code:

template<class T>
class Array2D
{
public:
    Array2D(int a, int b)  
    {
        num1 = (T**)new int [a*sizeof(int*)];
        for(int i = 0; i < a; i++)
            num1[i] = new int [b*sizeof(int)];

        for (int i = 0; i < a; i++) {
            for (int j = 0; j < b; j++) {
                num1[i][j] = i*j;
            }
        }
    }
    class Array1D
    {
    public:
        Array1D(int* a):temp(a) {}
        T& operator[](int a)
        {
            return temp[a];
        }
        T* temp;
    };

    T** num1;
    Array1D operator[] (int a)
    {
        return Array1D(num1[a]);
    }
};


int _tmain(int argc, _TCHAR* argv[])
{
    Array2D<int> arr(20, 30);

    std::cout << arr[2][3];
    getchar();
    return 0;
}

Answer 4

struct test
{
    using array_reference = int(&)[32][32];

    array_reference operator [] (std::size_t index)
    {
        return m_data[index];
    }

private:

    int m_data[32][32][32];
};

Found my own simple solution to this.

Answer 5

My 5 cents.

I intuitively knew I need to do lots of boilerplate code.

This is why, instead of operator [], I did overloaded operator (int, int). Then in final result, instead of m[1][2], I did m(1,2)

I know it is DIFFERENT thing, but is still very intuitive and looks like mathematical script.

Answer 6

You can use a proxy object, something like this:

#include <iostream>

struct Object
{
    struct Proxy
    {
        Object *mObj;
        int mI;

        Proxy(Object *obj, int i)
        : mObj(obj), mI(i)
        {
        }

        int operator[](int j)
        {
            return mI * j;
        }
    };

    Proxy operator[](int i)
    {
        return Proxy(this, i);
    }
};

int main()
{
    Object o;
    std::cout << o[2][3] << std::endl;
}