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

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]));

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