allocating memory to an array of variable length

Question

I want the length of this array to be taken as input from user. let n be the entered length; when I use malloc: int *arr1[n] = malloc((sizeof(int) * n) + 1); it says a variable sized array can't be initialized, is there any other way to do it?

I want to take an array as input from the user, sort the elements of the array and store the sorted elements in another array. Here's my complete code:

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

void swap(int *a, int *b)
{    
    int temp = 0;
    temp = *a;
    *a = *b;
    *b = temp;
}

int main(void)
{    
    int n = 0;
    printf("Enter the number of elements - ");
    scanf("%i", &n);
    int *arr1[n];
    int arr2[] = malloc((sizeof(int) * n) + 1);
    printf("Enter the elements one by one - ");
    for (int i = 0; i < n; i++)
    {
        scanf("%i", arr1[i]);
    }
    free(arr2);
}

I didn't declare the second array as arr2[n] as n is variable and it won't let me initialize the array with malloc. Can someone please help me with this?

int arr1[n] = malloc((sizeof(int) * n) + 1), why you make it as areay malloc just return generic pointer (pointer to void) location first address in the reserved block so just make it int *= (int)malloc((sizeof(int) * n) + 1); now you have an array !. — Looper
– Looper, Commented Apr 12, 2020 at 8:39
Why the + 1 ? And what is arr2 for? You don't use it at all? — Jabberwocky
– Jabberwocky, Commented Apr 12, 2020 at 8:40
make sure you're aware of the difference between %i and %d in scanf — M.M
– M.M, Commented Apr 12, 2020 at 8:56
This question has already been asked here: stackoverflow.com/questions/6634888/… — ghilesZ
– ghilesZ, Commented Apr 12, 2020 at 13:39

Jabberwocky · Accepted Answer · 2020-04-12 08:37:44Z

0

There is no other way to do it, but there is a correct way:

You want this:

int arr1[n];
int *arr2 = malloc((sizeof(int) * n) + 1);
...
scanf("%i", &arr1[i]);

answered Apr 12, 2020 at 8:37

Jabberwocky

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