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Home/ Questions/Q 8608353
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T03:35:26+00:00 2026-06-12T03:35:26+00:00

I’m following LCTHW tutorial and I have a task to do. This is the

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I’m following LCTHW tutorial and I have a task to do.
This is the data structure:

typedef struct DArray {
    int end;
    int max; 
    size_t element_size;
    size_t expand_rate;
    void **contents; 
} DArray;

I have declared a typedef:

typedef int (*DArray_compare) (const void *a, const void *b);

When I create a sorting function, I pass to it a DArray_compare, the problem is that I can’t figure out how to do an example of this comparator.

I tried to do something like this:

int compare(const void *a, const void *b)
{
  int i = (int)*a;
  int k = (int)*b;
  printf("%d %d\n", i, k);
  return i - k;
}

But I get an error:

error: operand of type 'void' where arithmetic or pointer type is required int i = (int)*a;

The question is: without changing the struct and the typedef of the comparator, I want to create a comparator that compares int, how can I do it?

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  1. Editorial Team
    Editorial Team
    2026-06-12T03:35:27+00:00Added an answer on June 12, 2026 at 3:35 am
    int i = *(int*)a;
    // This one has more parens to make it really obvious what your intent is.
    int k = *((int*)b);
    

    The second line (k=) is easiest to explain cos of all the brackets. You can rewrite it as follows:

    // Cast b from a pointer to a void into a pointer to an int.
    int *X = (int*)b; 
    // k = "what X is pointing to" or "the contents of X"
    int k = *X;
    

    edit:
    I think ralu’s comment is suggesting you change all the void* to int* which is a much safer solution if you have that power.

    typedef int (*DArray_compare) (const int *a, const int *b);
    
    int compare(const int *a, const int *b)
    {
        int i = *a;
        int k = *b;
        ...
    
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