How to create multi-dimensional array with dynamic rows and columns

Question

I am trying to code a array with dynamic rows and columns (multi-dimensional).

This is what I've tried:

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

#define LEN(arr) ((int) (sizeof (arr) / sizeof (arr)[0]))

int main() {
    int size, q = 0;
    // get arr size from stdin
    scanf("%d", &size);
    int *arr[size];

    for (int i=0; i<size; i++) {
        int len, element = 0;
        // get length of column from stdin
        scanf("%d", &len);
        arr[i] = malloc(len * sizeof(int));
        for(int j=0; j<len; j++) {
            // get each column's element from stdin and append to arr
            scanf("%d", &element);
            arr[i][j] = element;
        }
    }
    for (int i=0; i< LEN(arr); i++)
        printf("%d\n", LEN(arr[i]));
}

stdin:

2
3 4 5 6
4 1 2 3 4

The first line of input is the size/amount of arrays to be stored in arr (2), the following lines begin with the size of the columns (3, 4) followed by the elements to store (4 5 6, 1 2 3 4) in each column.

stdout:

2
2

When I run this program with the output is 2 meaning each column's length is 2, this is unintended. I am seeking for a solution to output the correct column lengths. What am I doing wrong?

The intended stdout is:

3
4

dbush · Accepted Answer · 2020-06-29 01:48:04Z

1

arr is an array, so sizeof(arr) gives you the size in bytes of the array as expected.

arr[i] is not an array however. It is a pointer, specifically a int *. So sizeof(arr[i]) gives you the size in bytes of an int *.

When you allocate memory dynamically, you need to keep track of how much space was allocated. You can do this by maintaining a separate array with size size of each subarray.

int *arr[size], arr_len[size];
...
for (int i=0; i<size; i++) {
    ...
    arr[i] = malloc(len * sizeof(int));
    arr_len[i] = len;
    ...
}

for (int i=0; i< LEN(arr); i++)
    printf("%d\n", arr_len[i]));

answered Jun 29, 2020 at 1:48

dbush

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