函数内二维数组的动态分配（使用指针返回分配对象的地址）

发布于 2024-12-02 16:48:55 字数 425 浏览 0 评论 0原文

我想知道如何使用函数参数将指针传递给动态分配的数组。该函数应该分配数组 10x10（为了简单起见，跳过了检查）。这可能吗？我做错了什么？提前致谢。

int array_allocate2DArray ( int **array, unsigned int size_x, unsigned int size_y)
{
    array = malloc (size_x * sizeof(int *));

    for (int i = 0; i < size_x; i++)
        array[i] = malloc(size_y * sizeof(int));

    return 0;
}

int main()
{
    int **array;
    array_allocate2DArray (*&array, 10, 10);
}

原文

I'd /ike to know, how to pass pointers to dynamically allocated arrays using function arguments. This function is supposed to allocate array 10x10 (checks skipped for simplicity sake). Is this possible? What am i doing wrong? Thanks in advance.

int array_allocate2DArray ( int **array, unsigned int size_x, unsigned int size_y)
{
    array = malloc (size_x * sizeof(int *));

    for (int i = 0; i < size_x; i++)
        array[i] = malloc(size_y * sizeof(int));

    return 0;
}

int main()
{
    int **array;
    array_allocate2DArray (*&array, 10, 10);
}

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不乱于心 2024-12-09 16:48:55

当我面临类似的问题时，我遇到了这篇文章（我正在寻找一种在 C 中动态分配字符串数组的方法）。我更喜欢从函数返回数组指针。以下内容对我有用（我根据您的整数数组对其进行了调整）。我为每个值任意设置 99，这样我就可以在 main.c 中看到它们打印出来。

int **array_allocate2DArray(unsigned int size_x, unsigned int size_y)
{
        int i;
        int **arr;

        arr = malloc(size_x*(sizeof(int*)));

        for (i=0 ; i<size_x ; i++){
                arr[i] = malloc(size_y);
                *arr[i] = 99;
        }

        return arr;

}

int main(void)
{
        int i;
        int **arr;

        arr = array_allocate2DArray(10, 10);

        for (i=0 ; i<10 ; i++){
                printf("%d\n", *arr[i]);
                free(arr[i]);
        }
        free(arr);

        return 0;
}

I came across this post when I was facing a similar problem (I was looking for a way to dynamically allocate an array of strings in C). I prefer to return the array pointer from the function. The following worked for me (I adapted it for your array of integers). I arbitrarily set 99 for each value so I could see them printed out in main.

int **array_allocate2DArray(unsigned int size_x, unsigned int size_y)
{
        int i;
        int **arr;

        arr = malloc(size_x*(sizeof(int*)));

        for (i=0 ; i<size_x ; i++){
                arr[i] = malloc(size_y);
                *arr[i] = 99;
        }

        return arr;

}

int main(void)
{
        int i;
        int **arr;

        arr = array_allocate2DArray(10, 10);

        for (i=0 ; i<10 ; i++){
                printf("%d\n", *arr[i]);
                free(arr[i]);
        }
        free(arr);

        return 0;
}

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静若繁花 2024-12-09 16:48:55

尝试这样的操作：

int array_allocate2DArray (int ***p, unsigned int size_x, unsigned int size_y)
{
    int **array = malloc (size_x * sizeof (int *));

    for (int i = 0; i < size_x; i++)
        array[i] = malloc(size_y * sizeof(int));

    *p = array;
    return 0;
}


int **array;
array_allocate2DArray (&array, 10, 10);

我使用临时 p 来避免混淆。

Try something like this:

int array_allocate2DArray (int ***p, unsigned int size_x, unsigned int size_y)
{
    int **array = malloc (size_x * sizeof (int *));

    for (int i = 0; i < size_x; i++)
        array[i] = malloc(size_y * sizeof(int));

    *p = array;
    return 0;
}


int **array;
array_allocate2DArray (&array, 10, 10);

I used the temporary p to avoid confusion.

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