警告：return 来自不兼容的指针类型 [-Wincompatible-pointer-types]|

Question

我正在尝试从函数指针 return 一个数组，代码可以工作，但在 C 中显示“不兼容的指针类型”的警告。我想要 return 一个数组，它已经是动态分配的。谁能告诉我问题和解决方法

#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;
}

有人可以给我这个警告的解决方案吗？

Answer 1

尝试

#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;
}

这将解决原始警告的主要问题。

重要的是为您要返回的内容分配正确类型的内存。问题是当请求指向 unsigned short* 的指针时，您返回 int* 。不这样做对以后访问内存有影响，因为从返回的指针访问的内存布局与您分配内存序列的方式不匹配。

Answer 2

如果你想要 return 一个 unsigned short 的数组，你应该分配一个 unsigned short 的数组而不是 ints:

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 3

您的函数声明为 return unsigned short *，但您为 int 分配了 space 并尝试 return 和 int*。我假设您想将 unsigned shorts 存储在分配的内存中。
当您 return 一个指向动态分配内存的指针时，您应该始终将该指针分配给一个变量，以便您可以 free 分配的内存。

#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 4

您正在分配一个整数数组，但您正在 return 分配一个短裤数组（通常是整数大小的一半）。只是 return 指向 int 的指针，如：

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

或者分配一个短裤数组，如：

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

警告：return 来自不兼容的指针类型 [-Wincompatible-pointer-types]|

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

c

pointers

function-pointers