How to make a function that returns a 2-dimensional array in c++

Question

I need to make this function. The parameter suppose to be a random N*N array. And it should returns a (N-1)*(N-1) array. How can I do this?

int** SubMatrix(int** matrix){
    int** submatrix=new int*[size-1];
        ................................
        ................................
        return submatrix;
}

Is this code correct?

Besides, say in the main function, I already have this array

x={{1,2,3},{1,2,3},{1,2,3}};

How do I call the function?

Calling:

int y[2][2]=SubMatrix[x]

is wrong and

int** y=SubMatrix[x]

is also wrong...

Is N a known constant, or can change at runtime? In other words, does it have to handle any size input array? — davec, Nov 12 '12 at 19:21
You should have a look at c++ basic syntax. A function should be called like `int** y = SubMatrix(x)` — kosklain, Nov 12 '12 at 19:21
Is that function signature fixed? If you can change that signature, there are much better ways to do this. — Yakk - Adam Nevraumont, Nov 12 '12 at 19:34
ideally the size should be variable. But I don't know how to define it even if the size is fixed. You cannot declare like int f[][]() right? — user1819047, Nov 13 '12 at 02:02

score 0 · Answer 1 · edited May 23 '17 at 11:43

First, there is no way to keep track of an array's size in C++ (only the amount of memory allocated). So, it is necessary that the size of the array be passed as a parameter or be global. I make this assumption in the following declaration:

int* SubMatrix(int** matrix, int size){
    int* submatrix = new int[(size - 1) * (size - 1)];
    // Do stuff
    return submatrix;
}

Second, if your matrix is square it may be better to use the declaration described above where the size of the array allocated is (N-1)*(N-1) (See Kevin Loney's answer How do I declare a 2d array in C++ using new? regarding matrix declaration on a heap); thus, the array is accessible as a matrix. Accessing cells will, however, require a bit of arithmetic:

for (int i = 0; i < size - 1; ++i)
    for (int j = 0; j < size - 1; ++j)
        submatrix[i * (size -1) + j] // Element at [i][j]

Finally, the array returned by the function is now a matrix pointer:

int* matrix = SubMatrix(matrix, size);

The dimensions of the matrix returned are (size-1)*(size-1)

With respect to your example now:

x = {{1,2,3},{1,2,3},{1,2,3}};
int* y = SubMatrix(matrix, 3);
// y is now a matrix of size (size-1)*(size-1)

If you really (really) want a two-dimensional array:

int** SubMatrix(int** matrix, int size){
    int** submatrix = new int*[size - 1];
    for (int i = 0; i < size - 1; ++i)
        submatrix[i] = new int[size - 1]
    // Do stuff
    return submatrix;
}

but this way the return is a one dimensional array right? Is there any way I can make it 2 dimensional? — user1819047, Nov 13 '12 at 02:03
Sure, but it isn't necessarily "2 dimensional"; rather, it's called "jagged" because you have an array of array pointers which may be of arbitrary size. I've appended the code to my answer. — Gio Borje, Nov 13 '12 at 04:03

How to make a function that returns a 2-dimensional array in c++

1 Answers1