警告：從不兼容的指針類型返回 [-Wincompatible-pointer-types]|

Question

我正在嘗試從 function 指針返回一個數組，代碼工作但在 C 中顯示“指針類型不兼容”的警告。 我想返回一個數組，它已經是動態分配的。 有人可以告訴我問題和解決方案嗎

#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;
}

有人可以給我這個警告的解決方案嗎？

Answer 1

• 您的 function 被聲明為返回unsigned short *但您為int分配空間並嘗試返回int* 。 我假設您想將unsigned short存儲在分配的 memory 中。
• 當您返回指向動態分配的 memory 的指針時，您應該始終將該指針分配給一個變量，以便您可以free分配的 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

嘗試

#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;
}

這將解決原始警告的主要問題。

重要的是分配您返回的正確類型的 memory。 問題是當請求指向unsigned short*的指針時，您正在返回一個int* 。 不這樣做會對稍后訪問 memory 產生影響，因為從返回的指針訪問的 memory 布局與您分配 memory 序列的方式不匹配。

Answer 3

如果你想返回一個unsigned short數組，你應該分配一個unsigned short數組，而不是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

您正在分配一個整數數組，但您正在返回一個短褲數組（通常是整數大小的一半）。 只需返回一個指向int的指針，如下所示：

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

或分配一個短褲數組，如：

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