is that valid to use (int*)calloc(size-1,2)?

Question

I am making a program where user can edit array. I am new in programming.code is not finished yet.

Can I use calloc(size-1,2)? Is that valid ? Does it creates array size-1?

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

 int main()
 {

int *pointer1,*pointer2,size,i,choice,j=0,index;

     printf("Enter Elements of Arrays " );
     scanf("%d",&size);

      pointer1=(int*)calloc(size,2);

     for(i=0;i!=size;i++){
        scanf("%d",pointer1+i);
     }

    printf("Enter your choice \n" );
    scanf("%d",&choice);

      switch(choice) 
     {
     case 0:
         printf("Enter Index for Deletation ");
         scanf("%d",&index);
 /* I know that code is not finish. but calloc(size-1,2) is that valid or not ?*/
         pointer2=(int*)calloc(size-1,2);
         for(i=0;i!=size-1;i++){
            if(i!=index){
             *pointer2+i=*pointer1+i+j;
                    }
                    else{
                 j++;  
              }
             }
                  for(i=0;i<=size;i++)
                  {
              printf("\n%d",*pointer2+i);
            }

           break;

     default:
         printf("Error!");
     } 
  return 0;
 }

It might be but it is not the best way. Please read https://stackoverflow.com/questions/7097678/removing-elements-from-dynamic-arrays for better ideas — Antti Haapala -- Слава Україні, Sep 05 '18 at 10:03
Your `calloc` call seems assume that `sizeof(int)` is 2, which might not be correct and certainly is not portable. If `sizeof(int)` is in fact 2, your `calloc` call allocates space for `size-1` elements, but your loop accesses elements at index `size-1` and `size` which will be beyond the end of the memory allocated by `calloc`. — Ian Abbott, Sep 05 '18 at 10:19
Your loop is `for(i=0;i<=size;i++)` so you certainly don't want `size-1` elements in the array, you need `size+1` elements. Moreover `pointer1[i+j]` is likely to index beyond the bounds of the memory allocation although we don't see the allocation for `pointer1`. — Weather Vane, Sep 05 '18 at 10:20
That first loop will never finish as you have `i--` and `i++` cancelling each other out, so it'll get stuck when `i==index` — Chris Turner, Sep 05 '18 at 10:27

score 0 · Answer 1 · answered Sep 05 '18 at 10:06

0

Function signature for calloc is as below.

void * calloc( size_t num, size_t size );

Make sure that num > 0 and size > 0

answered Sep 05 '18 at 10:06

Mayur

2,583
16
28

thanks for ans! but can i do calloc(size-1,2);. user will enter size. – Owais Qureshi Sep 05 '18 at 10:08
@OwaisQureshi I suggest `pointer2 = calloc(size + 1, sizeof *pointer2);` – Weather Vane Sep 05 '18 at 10:26
@WeatherVane but i want to delete element! that's why calloc(size-1,sizeof(pointer2));. thank you. – Owais Qureshi Sep 05 '18 at 10:47
@OwaisQureshi your code is `for(i=0;i<=size;i++){ ... pointer2[i]= ... ` etc. In the final iteration you are indexing `pointer2[size]` and therefore there must be `size+1` elements. – Weather Vane Sep 05 '18 at 10:50

Mathieu · Accepted Answer · 2018-09-05T10:13:53.003

Since you do not tell us what is pointer2, it's difficult to answer.

But, if you write:

int *pointer2 = NULL;
size_t size = 10; 

if (size-1 > 0)
{
    pointer2 = calloc(size-1, sizeof *pointer);
}

The pointer2 will points to an portion of memory able to store size-1 integer (int) values.

In C, it is not necessary to cast [m|c|re]alloc results
Magic values, like the 2 in calloc(..., 2); are dangerous : if the type of pointer2 changed, the allocation could become invalid, or if the type size change over some architectures, it can become problematic.

is that valid to use (int*)calloc(size-1,2)?

2 Answers2