Dynamic Memory Allocation Of an Array in C

Question

I have been trying to write a simple code wherein the array is allocated dynamically. Every time I specify the side of the array as n(suppose 4) and proceed to type the given input, it takes exactly n+1(5 in this case) inputs from me but as the output, it prints n(4) elements.

Here's the main function I wrote:

int main() {
    int *arr, n;
    scanf("%d", &n); //n is the size
    arr = (int *)malloc(n*sizeof(int));
    for(int i=0; i<n; i++) {
        scanf("%d ", &arr[i]);
    }
    for(int i=0 ; i<n; i++) {
        printf("%d ", arr[i]);
    }
}

I've also tried doing the code by initializing i in the first loop as 1, and in that way it takes exactly n inputs but it gives a weird output, something like this:

7953616 1 2 3

`scanf("%d ", &arr[i]);` --> `scanf("%d", &arr[i]);`. Drop space. Certainly a dupe someplace. — chux - Reinstate Monica, Apr 22 '21 at 18:29
@chux-ReinstateMonica yes, it does give the correct output now but why is that space mattering so much? — slayk, Apr 22 '21 at 18:32
`" "` consumes 0 or more white-spaces. So it does not return until seeing a following non-white-space to know all white-space consumed. — chux - Reinstate Monica, Apr 22 '21 at 18:33
[C FAQ: Why does everyone say not to use scanf? What should I use instead?](http://c-faq.com/stdio/scanfprobs.html) In particular [12.17](http://c-faq.com/stdio/scanfhang.html). — Schwern, Apr 22 '21 at 18:37
@Schwern Thank you, [12.17](http://c-faq.com/stdio/scanfhang.html) was really helpful. — slayk, Apr 22 '21 at 18:46

score 0 · Accepted Answer · answered Apr 22 '21 at 18:42

in those two lines

    scanf("%d ", &arr[i]);
    printf("%d ", arr[i]);

you have %d instead of %d

also you need to use free() from stdlib library as after you finish using the pointer you need to free the memory in order to reuse it again otherwise this will happen

final code

#include <stdio.h>
#include <stdlib.h>
int main() {
 int *arr, n;
 scanf("%d", &n); //n is the size
 arr = (int *)malloc(n*sizeof(int));
 for(int i=0; i<n; i++) 
 scanf("%d", &arr[i]);
 for(int i=0 ; i<n; i++) 
 printf("%d", arr[i]);
 free(arr);
}

Thank you. @Saifeddine Ben Salem – slayk Apr 22 '21 at 18:49 — slayk, Apr 22 '21 at 18:49

Dynamic Memory Allocation Of an Array in C

1 Answers1