在C中交换没有memcpy的void *指针数组的项 [英] Swap items of void* pointer array without memcpy in C

问题描述

我正在编写一些学校项目，并且需要交换两项void *指针数组.我可以使用以下代码来做到这一点:

I am writing some school project, and I need to swap two items of void* pointer array. I can do this with something like following code:

void swap(void *base, int len, int width)
{
    void *p = malloc(width);

    memcpy(p,base,width);
    memcpy(base,(char*)base+width,width);
    memcpy((char*)base+width,p,width);

    free(p);
}

但是我需要不使用memcpy交换项目，仅使用malloc，realloc和free.那有可能吗?

But I need to swap items WITHOUT memcpy, only with malloc, realloc and free. Is that even possible?

谢谢

推荐答案

为什么不以这种方式交换?:

Why don't swap in this way?:

void swap(void *v[], int i, int j)
{
    void *temp;

    temp = v[i];
    v[i] = v[j];
    v[j] = temp;
}

与qsort一样(在数组中交换元素):

As qsort does (swaps elements within the array):

void sort(void *v[], int left, int right, int (*comp)(const void *, const void *))
{
    int i, last;

    if (left >= right) return;
    swap(v, left, (left + right) / 2);
    last = left;
    for (i = left + 1; i <= right; i++) {
        if ((*comp)(v[i], v[left]) < 0)
            swap(v, ++last, i);
    }
    swap(v, left, last);
    sort(v, left, last - 1, comp);
    sort(v, last + 1, right, comp);
}

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