about the array in C

Question

I have written a following code (see code comments for the question),

#include<stdio.h>
int main()
{
    int size;
    scanf("%d",&size);
    int arr[size];    /*is it a valid statement?*/
    for(int i=1;i<=size;i++)
    {
        scanf("%d",&arr[i]);
        printf("%d",arr[i]);
    }
    return 0;
}

Answer 1

The use of a non constant array size is valid C99 but not C90. There is an older gcc extension allowing it.

Note that taking advantage of this make it harder to check that memory allocation succeed. Using it with values provided by the user is probably not wise.

Answer 2

You cannot allocate a dynamic array in C90 like that. You will have to dynamically allocate it with malloc like this:

int* array = (int*) malloc(sizeof(int) * size);

You also index the arrays starting at 0 not 1, so the for loop should start like this:

for(int i = 0; i < size; i++)

If you use the malloc method, you will also need to free the memory after you're done with it:

free(array);