warning: return from incompatible pointer type [-Wincompatible-pointer-types]|

I am trying to return an array from the function pointer, the code work but shows a warning in C that "incompatible pointer type". I want to return an array and it is already dynamic allocated. Can somebody tell me the problem and the solution to it

#include <stdlib.h>
#include<stdio.h>
unsigned short *reverse_seq(unsigned short num)
{
    if(num==0) return NULL;
    int size=num+1;
    int* numbers=(int *) malloc(size*sizeof(int));
    for(int i=0;i<num;i++){
        numbers[i]=num-i;
    }
    for(int i=0;i<num;i++){
        printf("%d ",numbers[i]);
    }
    return numbers;
}
int main(void)
{
    int num=5;
    reverse_seq(num);
    return 0;
}

Can somebody give me the solution to this warning?

>Solution :

• Your function is declared to return unsigned short * but you allocate space for ints and try to return an int*. I assume you want to store unsigned shorts in the allocated memory.
• When you return a pointer to dynamically allocated memory, you should always assign that pointer to a variable so that you can free the allocated memory.

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

unsigned short *reverse_seq(unsigned short num) {
    if (num == 0) return NULL;

    // corrected allocation (there's no need for num + 1 elements either):
    unsigned short *numbers = malloc(num * sizeof *numbers);

    if(numbers) {                       // check that allocation worked
        for (int i = 0; i < num; i++) {
            numbers[i] = num - i;
        }
    }

    // printing moved to `main` to make use of the data there

    return numbers;
}

int main(void) {
    unsigned short num = 5; // same type as `reverse_seq` wants
    
    unsigned short *numbers = reverse_seq(num);
    if(numbers) {                        // again, check that allocation worked
        for (int i = 0; i < num; i++) {
            printf("%d ", numbers[i]);
        }
        
        free(numbers);                   // free the memory
    }
}