I am new to C language and I try to write function to allocate memory for 2d array What am I doing:

void allocate(int **arr, int r, int c) 
{
    **arr = (int **)malloc(r*c*sizeof(int));
}
    int main( void )
{
    int NO_OF_COLS = 0;
    int NO_OF_ROWS = 0;    
    scanf("%d%d", &NO_OF_ROWS, &NO_OF_COLS);

    int **matrix;
    
    allocate(matrix, NO_OF_ROWS, NO_OF_COLS);

    return 0;
}

I have this warning: assignment to 'int' from 'int **' makes integer from pointer without a cast [-Wint-conversion] 8 | **arr = (int **)malloc(rcsizeof(int)); | ^

I understand that I am passing memory to 'matrix' in allocate(), but I don't understand how I can return new memory address and assign it to matrix

I try to change allocate(matrix, NO_OF_ROWS, NO_OF_COLS); to allocate(&matrix, NO_OF_ROWS, NO_OF_COLS); but it still doesn't work

CodePudding user response：

First of all you need to allocate an array of arrays to hold the data. You need to allocate an array to hold N elements of pointer type. Consider this:

int** allocate(int row, int col)
{
    int i;
    int** a = (int**) malloc(row * sizeof(int*));
    if (!a)
        return  0;

    for(i=0; i < row; i  ) {
        a[i] = (int*) malloc(sizeof(int) * col);
    }
    return  a;
}

You need to allocate a to hold N elements that are pointers to int arrays, then loop and allocate the actual array needed ( sizeof(int) * col ) in your use case. So here is in use:

int main()
{
    int** a= allocate(10, 10);
    int i=0, j=0;
    for(i=0; i < 10; i  ) {
        for(j=0; j < 10; j  ) {
            a[i][j] = 42;
        }
    }

    for(i=0; i < 10; i  ) {
        for(j=0; j < 10; j  ) {
            printf("[%d]", a[i][j]);
        }
        puts("");
    }
    return 0;
}

Also you need to write a deallocation function the same way, but first you have to loop trough all elements and free the arrays before freeing the a array in your case. [EDIT]: I haven't added a fail safe check to the second malloc for simplicity. Given that example and that function prototype you need to allocate a 1D array as:

void allocate(int** a, int row, int col)
{
    *a = malloc(sizeof(int) * row * col);
}

But it's a different than allocating arrays of arrays.

CodePudding user response：

Since you can't modify the parameters for allocate, you will need to manually compute the index of each element in the matrix. The following bare-bones program shows how to access each element of a 5 x 5 matrix.

#include <stdio.h>
#include <stdlib.h>

void allocate(int **arr, int r, int c)
{
        *arr = malloc(sizeof (int) * r * c); 
}

int main(void)
{
        int *arr = NULL;
        int NO_OF_ROWS = 5, NO_OF_COLS = 5;
        allocate(&arr, NO_OF_ROWS, NO_OF_COLS);

        for (int r = 0; r < NO_OF_ROWS;   r) {
              for (int c = 0; c < NO_OF_COLS;   c) {
                    printf("arr[%d]\n", r * NO_OF_COLS   c); 
              }   
        }   
}

Here, we are actually using a 1-D array. Inside the loop, r * NO_OF_COLS c is responsible for accessing each element. This program will print arr[0], arr[1]...arr[24]. You may use this logic to calculate the index of each element.

Inside main(), arr is just a pointer to an int. When we pass the address of arr to allocate(), we are giving allocate() a chance to modify arr. Then, inside allocate(), we hop to the address that was passed in and lay down the bits returned by malloc.

CodePudding user response：

Here is a simpler form of what you try to do:

void compute_with_matrix(int NO_OF_ROWS, int NO_OF_COLS) 
{
    int matrix[NO_OF_ROWS][NO_OF_COLS];

    /* here work with allocated matrix */  
}
    int main( void )
{
    int NO_OF_COLS, NO_OF_ROWS;    
    scanf("%d %d", &NO_OF_ROWS, &NO_OF_COLS);
    compute_with_matrix(NO_OF_ROWS, NO_OF_COLS);
    return 0;
}