Returning a multidimensional array in C++

Question

void createArray(int a, int b, int c, int d, int array[3][3]){

    int state[3][3];

    for(int x=0;x<3;x++){
        for(int y=0;y<3;y++){

            if(x == a && y == b){
                state[x][y] = array[c][d];
            }
            else if(x == c && y == d){
                state[x][y] = array[a][b];
            }
            else{
                state[x][y] = array[x][y];
            }
        }
    }

    for(int i=0;i<3;i++){
        for(int j=0;j<3;j++){
            cout << state[i][j] << " ";
        }
        cout << endl;
    }

}

I have basically got this function which clones the multidimensional array that is inputed but swaps the values of the two co-ordinates (a,b) and (c,d) around. This is then outputted out to the console.

However what I would really like is for this to be returned as a multidimensional array, but I don't think this can be done?

I have looked at vectors and pointers but don't really understand them and if I use them, I will then have to change all the previous code I have written.

@πάνταῥεῖ I don't think there is a need for MCVE here. The question is simple: returning a 2D array from a function. — bolov
– bolov, Commented Dec 7, 2014 at 20:01
@bolov Where's the return actually? What's the actual inputs/outputs/errors? — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Dec 7, 2014 at 20:02
@πάνταῥεῖ array is the input 2D array. state is the 2D array he wants to return. — bolov
– bolov, Commented Dec 7, 2014 at 20:04

Peter · Accepted Answer · 2014-12-07 20:31:23Z

3

There are multiple options.

Option 1 - pass the array to the function

void createArray(int a, int b, int c, int d, const int array[3][3], int outArray[3][3]){
...
}

Option 2 - return a reference to an array - just make sure the array does not live on the stack of the function it's returned from.

typedef int My3Times3Array[3][3];
const My3Times3Array& createArray(int a, int b, int c, int d, const int array[3][3]){
...
}

Option 3 - return a class that contains an array

struct Array
{
    int array[3][3];    
};

...

Array createArray(int a, int b, int c, int d, const int array[3][3]){
...
}

There's also std::vector, std::array, or boost::matrix, but since you mentioned you aren't comfortable with pointers yet, I'd save template classes for later.

edited Dec 7, 2014 at 20:31

answered Dec 7, 2014 at 20:19

Peter

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Comments

anatolyg · Accepted Answer · 2014-12-08 09:11:57Z

3

When you want to return a non conventional data type (int, char etc), the best way of doing it is by making your very own one.

struct mat3
{
    int myArray[3][3];    
};


mat3 createArray(int a, int b, int c, int d, int array[3][3]){

mat3 state;

for(int x=0;x<3;x++){
    for(int y=0;y<3;y++){

        if(x == a && y == b){
            state.myArray[x][y] = array[c][d];
        }
        else if(x == c && y == d){
            state.myArray[x][y] = array[a][b];
        }
        else{
            state.myArray[x][y] = array[x][y];
        }
    }
}

for(int i=0;i<3;i++){
    for(int j=0;j<3;j++){
        cout << state.myArray[i][j] << " ";
    }
    cout << endl;
    return state;
}

}

I have looked at vectors and pointers but don't really understand them and if I use them, I will then have to change all the previous code I have written

I suggest you study pointers further, they are so essential that you are already using them without knowing it.

edited Dec 8, 2014 at 9:11

anatolyg

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answered Dec 7, 2014 at 20:07

senex

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2 Comments

anatolyg Over a year ago

"by making your very own" - looks like a typo. What does this mean?

senex Over a year ago

Make your own data type. In this case you are making a type called 'mat3'.

Vlad from Moscow · Accepted Answer · 2014-12-07 20:29:15Z

0

You can dynamically allocate a two-dimensional array and return pointer to it. For example

typedef int ( *STATE )[3];

STATE createArray(int a, int b, int c, int d, int array[3][3]){

    STATE state = new int[3][3];

    //...

    return state;
}

Or

int ( *createArray(int a, int b, int c, int d, int array[3][3] ) )[3]{

    int ( * state )[3] = new int[3][3];

    //...

    return state;
}

In fact the same result you can get if you will use vectors. For example

#include <vector>
#include <array>

//...
std::vector<std::array<int, 3>> createArray(int a, int b, int c, int d, int array[3][3]){

    std::vector<std::array<int, 3>> state( 3 );

    //...

    return state;
}

edited Dec 7, 2014 at 20:29

answered Dec 7, 2014 at 20:07

Vlad from Moscow

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4 Comments

anatolyg Over a year ago

In c++, using pointers is usually not advised. Also, std::vector<std::array<int, 3>> is sufficiently different from int[3][3] to make some additional explanation necessary.

AnArrayOfFunctions Over a year ago

Which is stupid and non-sense waste of program time, as the array size is already known.

Vlad from Moscow Over a year ago

@Jako I showed the general approach how arrays are allocated and how to work with pointers. It is not a stupid approach as you think. It is a flexible approach that will work with arrays of any size.

anatolyg Over a year ago

@VladfromMoscow Would it be better to use typedef int STATE[3] and have the function return STATE*? I guess it's functionally equivalent (not sure) but more readable.

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