警告：从不兼容的指针类型返回 [-Wincompatible-pointer-types]|

Question

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?

Answer 1

• Your function is declared to return unsigned short * but you allocate space for int s and try to return an int* . I assume you want to store unsigned short s 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
    }
}

Answer 2

Try

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

unsigned short *reverse_seq(unsigned short num) {
    if (num == 0) return NULL;
    int size = num + 1;
    unsigned short *numbers = malloc((size_t) size * sizeof(unsigned short));  // THIS IS THE IMPORTANT CHANGE
    for (int i = 0; i < num; i++) {
        numbers[i] = (unsigned short) (num - i);
    }
    for (int i = 0; i < num; i++) {
        printf("%d ", numbers[i]);
    }
    return numbers;
}

int main(void) {
    int num = 5;
    unsigned short *res = reverse_seq((unsigned short) num);
    free(res);
    return 0;
}

This will fix your major problem with the original warning.

The important thing is to allocate memory of the correct type of what you are returning. The problem is that you are returning a int* when a pointer to unsigned short* is requested. Not doing this has consequences for accessing the memory later on since the memory layout of the access from the returned pointer does not match how you allocated your memory sequence.

Answer 3

If you want to return an array of unsigned short you should allocate an array of unsigned short but not of int s:

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

    unsigned short* numbers = malloc(num * sizeof(unsigned short));
    if (numbers) {
        for (unsigned short i = 0; i < num; i++) {
            numbers[i] = num - i;
        }
    }
    return numbers;
}

Answer 4

You are allocating an array of integers, but you are returning an array of shorts (which normally are half the size of an integer). Just return a pointer to int , as in:

int *reverse_seq(unsigned short num) {
    /* ... */

or allocate an array of shorts, as in:

    unsigned short *numbers = malloc(num * sizeof *numbers);  /* and please, ***never*** cast the result of malloc() */