Get pointer to column of 2D array in C

Question

I am new to C programming and especially to pointers. In the program I wrote, I tried to write a function that returns a pointer to specified column of array. See the code below for better understanding (or confusion :) ):

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

// function for getting pointer to specidifed column index
// 'ind' is index of requested column, 'ncol' is number of items in column
int* get_col(const int* arr, unsigned int ind, unsigned int ncol);

int main() {
    unsigned int n;

    printf("Input matrix size : ");
    scanf("%i", &n);

    int arr[n][n];

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            arr[i][j] = i * n + j;
    }

    for (int i = 0; i < n; i++) {
        printf("values in column %d: \n", i);
        int *col = get_col((int*)arr, i, n);
        for (int j = 0; j < n; j++) {
            printf("%d ", *col);
            col = col + 1;
        }
        printf("\n");
    }

    return 0;
}

int* get_col(const int* arr, unsigned int ind, unsigned int ncol) {
    int *result = malloc(sizeof(int) * ncol);

    for (int i = 0; i < ncol; i++)
        *result = *(arr + i*ncol + ind);

    return result;
}

As you see get_col function accepts pointer to array, column index and column size (n of elements in column, i.e number of rows) as arguments and trying to return a pointer to 1D array that contains values of column at requested index. The problem is that result is not correct. In case n=3 results are like below:

Input matrix size : 3
values in column 0: 
6 0 0 // supposed to be 0 3 6
values in column 1: 
7 0 0 // supposed to be 1 4 7
values in column 2: 
8 0 0 // supposed to be 2 5 8

I think that the problem lies in my understanding of pointers not the algorithm implemented. Actually, at first I didn't use pointer in my get_col function like below:

int result[ncol];

// ... do my work here to populate array

return &result;

Then as compiler complains warning: function returns address of local variable [-Wreturn-local-addr], I converted result from array to pointer in get_col function like above. What is the problem in this code? Did I use pointers in get_col function as it should be?

You already have a nice answer, but remember to free the pointer that you allocated inside the "get_col" function before the program finishes. In this case is not very problematic because the program ends and the operative system will do it, but it's a good habit to free all the memory you allocate, otherwise your programs will run full of memory leaks, and then crashes because "out of memory". it's just a tip since you say you're new to C and pointers. ;) — Raúl Herrero
– Raúl Herrero, Commented Apr 17, 2020 at 17:52
@RaúlHerrero thanks for the tip. Can you tell exactly where i should call free in this code? — Elgin Cahangirov
– Elgin Cahangirov, Commented Apr 17, 2020 at 17:56
free any pointer after you are not going to use it anymore. In this case, after the last printf, at the end of every loop of the for loop (second i++ loop, which it's used for printing the results), where you still have the pointer defined. — Raúl Herrero
– Raúl Herrero, Commented Apr 18, 2020 at 19:14

Youssef13 · Accepted Answer · 2020-04-17 17:33:50Z

1

In the following line:

*result = *(arr + i*ncol + ind);

You're always writing to the same memory address.

Change it to one of the two following options:

*(result + i) = *(arr + i*ncol + ind);
result[i] = *(arr + i*ncol + ind);

Regarding your second problem when you used:

int result[ncol];

// ... do my work here to populate array

return &result;

You should understand that result variable in this case (static-memory allocation) is stored in the stack. So, after your function returns, the variable values doesn't exist anymore in the memory. That's why you need dynamic-memory allocation. In dynamic-memory allocation, that values stay in the memory until you call free by yourself.

edited Apr 17, 2020 at 17:33

answered Apr 17, 2020 at 17:27

Youssef13

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