Using Pointers in lieu of Arrays

Question

I am definitely new to the C game, and would love some help with the following code snippet:

#include <stdio.h>

int main() {
    int cases;
    scanf("%d", &cases);
    printf("%d", cases);

    int i;
    int *heights;
    for(i=0; i<cases; i++){
        scanf("%d", &heights[i]);
    }

    return 0;
}

I understand that it segfaults because I'm giving scanf a NULL pointer, so is there any way to allow scanf to feed values into this pointer? Or is there a better method to get a variable number of arguments from stdin which I'm missing entirely?

On a side note you are (probably) not giving anything a NULL pointer. Leaving variables unassigned means they have an undefined value so could be anything. — DrYap
– DrYap, Commented Jun 19, 2014 at 8:24
The main problem is heights is defined as a pointer and you have not (probably via malloc()) given it anything to point to. — user3629249
– user3629249, Commented Jun 20, 2014 at 5:34

haccks · Accepted Answer · 2014-06-19 08:42:01Z

7

Use malloc to dynamically allocate space for heights.

int *heights = malloc(cases*sizeof(int));

And free pointer by calling free(heights) when you are done with heights.

For small value of cases you can use variable length arrays:

 int heights[cases];

and do not forget to compile your code in C99 mode (-std=c99).

edited Jun 19, 2014 at 8:42

answered Jun 19, 2014 at 8:18

haccks

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1 Comment

WhozCraig Over a year ago

+1 And make a habit of invoking free() for anything you malloc().

ayusha · Accepted Answer · 2014-06-20 05:49:09Z

3

heights pointer does not contain any memory.so first you have to allocate the memory using malloc system call according to the requirement. syntax for malloc in your case -

heights = (int *)malloc(cases*sizeof(int));

one thing to remember, after dynamic memeory location you have to free it.syntax for memory free -

free(heights)

edited Jun 20, 2014 at 5:49

answered Jun 19, 2014 at 8:36

ayusha

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