在初始化函数数组 [英] Initialize array in function

问题描述

当我想通过函数初始化一个指向数组的指针，我做以下内容：

When I want to initialize a pointer to an array through the function, I am doing the following:

初​​始化，并通过功能破坏阵：

Initialize and destroy array through functions:

int initArr(int **b)
{
    int *arr = (int *) malloc(sizeof(int)*2);

    if(arr == NULL)
        return 0;

    *b = arr;
    arr = NULL;
    return 1;
}

void destroyArr(int *b)
{
    free(b); 
    b = NULL;
}

初​​始化指针数组：

Initialize pointer to array:

int *pArr;
int initStatus = initArr(&pArr);

if(initStatus == 0)
{
    printf("%s", "error");
    return 0;
}

与指针数组工作：

Working with pointer to array:

*pArr = 1;
*(pArr + 1) = 2;

printf("0 = %i\n", *pArr);
printf("1 = %i\n", *(pArr + 1));

销毁指针数组：

Destroy pointer to array:

destroyArr(pArr);
pArr = NULL;

这是正确的和安全的？

Is this correct and safe?

推荐答案

我没有测试它，但它似乎是正确的。 A小调评论，虽然：你不需要设置改编或 B 到 NULL ，他们在其范围的尽头，不能（安全）之后访问的反正。

I haven't tested it, but it appears correct. A minor comment, though: you don't need to set arr or b to NULL, they're at the very end of their scope and can't be (safely) accessed after that anyway.

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