Return a 2d array from a function - c++

Question

Hi I am a newbie to C++
I am trying to return a 2d array from a function.
It is something like this
int **MakeGridOfCounts(int Grid[][6])
{
int cGrid[6][6] = {{0, }, {0, }, {0, }, {0, }, {0, }, {0, }};
return cGrid;
}

Answer 1

This code returns a 2d array.
#include <cstdio>
// Returns a pointer to a newly created 2d array the array2D has size [height x width]
int** create2DArray(unsigned height, unsigned width)
{
int** array2D = 0;
array2D = new int*[height];
for (int h = 0; h < height; h++)
{
array2D[h] = new int[width];
for (int w = 0; w < width; w++)
{
// fill in some initial values
// (filling in zeros would be more logic, but this is just for the example)
array2D[h][w] = w + width * h;
}
}
return array2D;
}
int main()
{
printf("Creating a 2D array2D\n");
printf("\n");
int height = 15;
int width = 10;
int** my2DArray = create2DArray(height, width);
printf("Array sized [%i,%i] created.\n\n", height, width);
// print contents of the array2D
printf("Array contents: \n");
for (int h = 0; h < height; h++)
{
for (int w = 0; w < width; w++)
{
printf("%i,", my2DArray[h][w]);
}
printf("\n");
}
// important: clean up memory
printf("\n");
printf("Cleaning up memory...\n");
for (int h = 0; h < height; h++) // loop variable wasn't declared
{
delete [] my2DArray[h];
}
delete [] my2DArray;
my2DArray = 0;
printf("Ready.\n");
return 0;
}

Answer 2

A better alternative to using pointers to pointers is to use std::vector. That takes care of the details of memory allocation and deallocation.
std::vector<std::vector<int>> create2DArray(unsigned height, unsigned width)
{
return std::vector<std::vector<int>>(height, std::vector<int>(width, 0));
}

Answer 3

That code isn't going to work, and it's not going to help you learn proper C++ if we fix it. It's better if you do something different. Raw arrays (especially multi-dimensional arrays) are difficult to pass correctly to and from functions. I think you'll be much better off starting with an object that represents an array but can be safely copied. Look up the documentation for std::vector.
In your code, you could use vector<vector<int> > or you could simulate a 2-D array with a 36-element vector<int>.

Answer 4

What you are (trying to do)/doing in your snippet is to return a local variable from the function, which is not at all recommended - nor is it allowed according to the standard.
If you'd like to create a int[6][6] from your function you'll either have to allocate memory for it on the free-store (ie. using new T/malloc or similar function), or pass in an already allocated piece of memory to MakeGridOfCounts.

Answer 5

The function returns a static 2D array
const int N = 6;
int (*(MakeGridOfCounts)())[N] {
static int cGrid[N][N] = {{0, }, {0, }, {0, }, {0, }, {0, }, {0, }};
return cGrid;
}
int main() {
int (*arr)[N];
arr = MakeGridOfCounts();
}
You need to make the array static since it will be having a block scope, when the function call ends, the array will be created and destroyed. Static scope variables last till the end of program.

Answer 6

#include <iostream>
using namespace std ;
typedef int (*Type)[3][3] ;
Type Demo_function( Type ); //prototype
int main (){
cout << "\t\t!!!!!Passing and returning 2D array from function!!!!!\n"
int array[3][3] ;
Type recieve , ptr = &array;
recieve = Demo_function( ptr ) ;
for ( int i = 0 ; i < 3 ; i ++ ){
for ( int j = 0 ; j < 3 ; j ++ ){
cout << (*recieve)[i][j] << " " ;
}
cout << endl ;
}
return 0 ;
}
Type Demo_function( Type array ){/*function definition */
cout << "Enter values : \n" ;
for (int i =0 ; i < 3 ; i ++)
for ( int j = 0 ; j < 3 ; j ++ )
cin >> (*array)[i][j] ;
return array ;
}

Answer 7

Whatever changes you would make in function will persist.So there is no need to return anything.You can pass 2d array and change it whenever you will like.
void MakeGridOfCounts(int Grid[][6])
{
cGrid[6][6] = {{0, }, {0, }, {0, }, {0, }, {0, }, {0, }};
}
or
void MakeGridOfCounts(int Grid[][6],int answerArray[][6])
{
....//do the changes in the array as you like they will reflect in main...
}

Answer 8

int** create2DArray(unsigned height, unsigned width)
{
int** array2D = 0;
array2D = new int*[height];
for (int h = 0; h < height; h++)
{
array2D[h] = new int[width];
for (int w = 0; w < width; w++)
{
// fill in some initial values
// (filling in zeros would be more logic, but this is just for the example)
array2D[h][w] = w + width * h;
}
}
return array2D;
}
int main ()
{
printf("Creating a 2D array2D\n");
printf("\n");
int height = 15;
int width = 10;
int** my2DArray = create2DArray(height, width);
printf("Array sized [%i,%i] created.\n\n", height, width);
// print contents of the array2D
printf("Array contents: \n");
for (int h = 0; h < height; h++)
{
for (int w = 0; w < width; w++)
{
printf("%i,", my2DArray[h][w]);
}
printf("\n");
}
return 0;
}

Answer 9

returning an array of pointers pointing to starting elements of all rows is the only decent way of returning 2d array.

Answer 10

I would suggest you Matrix library as an open source tool for c++, its usage is like arrays in c++. Here you can see documention.
Matrix funcionName(){
Matrix<int> arr(2, 2);
arr[0][0] = 5;
arr[0][1] = 10;
arr[1][0] = 0;
arr[1][1] = 44;
return arr;
}