How use a dynamic array with a structure

Question

I have written a program for heap sort and I tried to have both the array and the size of the array in a single structure but there seems to be some error in my code (I think it's in the initializing part). Here is my code.

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

int cmp;

typedef struct heap
{
    int *arr, n;
}heap;

void main()
{
    heap *h;
    scanf("%d",&h->n);
    h->arr = (int*)malloc(h->n*sizeof(int));
    
    for(int i=0; i<h->n; i++)
        scanf("%d",h->arr+i);
    
    build_heap(h);
    heap_sort(h);
    
    for(int i=0; i<h->n; i++)
        printf("%d ",h->arr[i]);
    printf("\n%d\n",cmp);
    free(h->arr);

}

I compiled it in linux platform and the error I got was Segmentation fault (core dumped) Can someone explain why I got this error and give a possible solution to this.

Using h->n and h->arr directly after declaring h as an uninitialised heap * cannot be correct. Instead, declare h as a heap and use h.n and h.arr. Remember that h->n is nothing more than a synonym for (*h).n, and using the * operator on an uninitialised pointer will result in a segmentation fault at best. — Stef
– Stef, Commented Oct 12, 2020 at 15:11

Vlad Rusu · Accepted Answer · 2020-10-12 14:52:02Z

3

By doing this heap *h; you have declared a pointer to a heap struct, but what does this pointer actually point to at this point?

You only need to allocate the memory for your heap:

void main()
{
    heap *h;
    // Alloc and initialize the pointer to the heap
    h = malloc(sizeof(heap));
    scanf("%d",&h->n);
    h->arr = malloc(h->n*sizeof(int));
    
    for(int i=0; i<h->n; i++)
        scanf("%d",h->arr+i);
    
    build_heap(h);
    heap_sort(h);
    
    for(int i=0; i<h->n; i++)
        printf("%d ",h->arr[i]);
    printf("\n%d\n",cmp);
    free(h->arr);

    // Don't forget to free the heap as well!
    free(h);

}

Also, it is a good practice to not cast the return value of malloc.

answered Oct 12, 2020 at 14:52

Vlad Rusu

1,48912 silver badges17 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

0___________ · Accepted Answer · 2020-10-12 14:58:33Z

1

Do it a bit different way:

typedef struct heap
{
    size_t n;
    int arr[]; 
}heap;

only one malloc and free is needed.

    size_t size;
    heap *h;

    if(scanf("%zu", &size) != 1) { /* error handling*/};
    h = malloc(sizeof(*h) + size * sizeof(h -> arr[0]));
    if(!h) { /* error handling*/};
    h -> size = size;

    /*...... */


    free(h);

answered Oct 12, 2020 at 14:58

0___________

71.6k4 gold badges41 silver badges94 bronze badges

Comments

Jabberwocky · Accepted Answer · 2020-10-12 15:05:21Z

1

When you declare heap *h; you declare a pointer, but after you never initialize h to point anywhere.

You probably want this:

int main()
{
  heap h;                              // now h is a heap and no more a pointer to heap
  scanf("%d", &h.n);
  h.arr = malloc(h.n * sizeof(int));   // the cast to (int*) is useless

  for (int i = 0; i < h.n; i++)
    scanf("%d", h.arr + i);            //BTW: &h.arr[i] would be more readable

  build_heap(&h);
  heap_sort(&h);

  for (int i = 0; i < h.n; i++)
    printf("%d ", h.arr[i]);

  printf("\n%d\n", cmp);
  free(h.arr);
}

There may be more problems in build_heap and in build_heap, but I can't tell you more as you didn't show these functions.

answered Oct 12, 2020 at 15:05

Jabberwocky

51.3k18 gold badges71 silver badges127 bronze badges

Collectives™ on Stack Overflow

How use a dynamic array with a structure

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related