Put a 2D array into existed memory using double pointer in C

Question

totally a newbie in C. I'm trying to put a 2D array into an existing memory space created by malloc. Here is the code:

int main()
{
    int **a;  //double pointer
    int i;
    void *ptr = malloc(sizeof(int)*4);  //2x2 array
    a = (int **)ptr;  //start of the array
    for(i=0; i<2; i++)  
            a[i] = (int *)a + i*2;
    printf("a: %p\n", a);
    printf("a[0]: %p\n", a[0]);
    printf("a[1]: %p\n", a[1]);
}

output:

a: 0x1976010
a[0]: 0x1976010
a[1]: 0x1976018

but when I try to access the element:

for(i=0; i<2; i++)
    for(j=0; j<2; j++)
        a[i][j] = j + i*10;

I got segment fault. Where did I do wrong? Thanks in advance!

Answer 1

If you have an existing (aka already allocated) memory area that you want to use for a 2D array (or rather for an array of array), you should not use a double pointer. Instead you should use a pointer to array of size N

Like

int (*a)[2];  // declare a as a pointer to array (size 2) of int

A complete program could be:

int main()
{
    // Allocate a memory area for holding 4 int
    void *ptr = malloc(sizeof(int)*4);

    int (*a)[2];  // a as a pointer to array (size 2) of int
    a = ptr;      // make a point to your memory area

    // Now a can be used as a "2D array" and data will be stored in the malloc'ed memory
    for (int i=0; i<2; ++i)
        for (int j=0; j<2; ++j)
            a[i][j] = i*100 + j;

    for (int i=0; i<2; ++i)
        for (int j=0; j<2; ++j)
            printf("a[%d][%d]=%d\n", i, j, a[i][j]);

    free(ptr);
    return 0;
}

Output:

a[0][0]=0
a[0][1]=1
a[1][0]=100
a[1][1]=101

Answer 2

When it comes to dynamic 2D arrays, there's unfortunately a lot of myths and hogwash, widely used bad practice and so on. Forget about pointer-to-pointers, is needlessly complex, error prone and slow.

A detailed explanation of the correct way to do this can be found here: Correctly allocating multi-dimensional arrays.

The correct way to allocate a 2D array dynamically is this:

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

int main()
{
    const size_t X = 2;
    const size_t Y = 2;

    int (*a)[Y] = malloc( sizeof(int[X][Y]) );
    int count = 1;

    for(size_t x=0; x<X; x++)
    {
      for(size_t y=0; y<Y; y++)
      {
        a[x][y] = count++; // assign some value here
        printf("%d ", a[x][y]);
      }
      printf("\n");
    }

    free(a);
}

Where int (*a)[Y] is an array pointer to the first element of the 2D array. The first element being a 1D array of type int [Y].

Answer 3

You got segmentation fault because a (a double pointer) is actually having a single pointer value.

There are multiple ways to use 2D array. The following is one of them.

#define ROW (2)
#define COL (3)

int main()
{
    int (*a)[COL] = malloc(sizeof(int[ROW][COL]));
    if(NULL == a) return -1; //Check if allocation is successfull.
    int i, j;
    for(i=0; i<ROW; i++)  
       for(j=0; j< COL; j++)
            a[i][j] = i*ROW + j; //Assign some values.
    printf("a: %p\n", a);
    printf("a[0][0]: %d\n", a[0][0]); //Printing a value.

    free(a); //Free the allocated memory.
}

Answer 4

EDIT

This isn't a correct answer, as pointed out by Lundin. There are much better ways to do this I wasn't aware of!

This line

a[i][j] = j + i*10;

is treating a[0] and a[1] as pointers themselves to other arrays, so it's reading their values

a[0]: 0x1976010
a[1]: 0x1976018

as memory locations, and segfaulting.

If you want to dynamically allocate 2d arrays, allocate one dimension of int*s, then iterate through it and allocate memory to each of the indices as ints.

int main()
{
    #define DIM_X 2
    #define DIM_Y DIM_X

    int i = 0;
    int j = 0;
    int** x = 0;
    void* a = malloc(sizeof(int*)*DIM_X);

    x = (int**)a;
    for (i = 0; i < 2; i++)
    {
        x[i] = (int*)malloc(sizeof(int)*DIM_Y);
    }

    // ...

    for (i = 0; i < DIM_X; i++)
    {
        free(x[i]);
    }
    free(x);

    return 0;
}

You could also do some tricky logic to keep everything contiguous, but it may be harder to read.

for (i = 0, j = 0; i < 2; i++)
{
    a[(i * 2) + j] = j + (i * 10);

    j++;
    if (2 < j) j = 0;
}

Answer 5

You have to allocate also storage for the pointers.

int main()
{
    int **a;  //double pointer
    int i;
    void *ptr = malloc(sizeof(int *)*2+sizeof(int)*4);  //2x2 array
                            //plus 2xpointers
    a = (int **)ptr;  //start of the array
    for(i=0; i<2; i++)  
            a[i] = (int *)a + 2 + i * 2; //plus 2 because the first 2 elements
                    //are your pointers
    printf("a: %p\n", a);
    printf("a[0]: %p\n", a[0]);
    printf("a[1]: %p\n", a[1]);
}