复制较小的阵列到阵列较大 [英] copy smaller array into larger array

问题描述

我有字符，分配了两个数组如下：

I have two arrays of chars, allocated as follows:

 unsigned char *arr1 = (unsigned char *)malloc((1024*1024) * sizeof(char));
 unsigned char *arr2 = (unsigned char *)malloc((768*768) * sizeof(char));

我想ARR2到ARR1复制，但preserve行/列结构。这意味着，只在第一个768个字节每个第一768行的将在ARR1被改变。

I would like to copy arr2 into arr1, but preserve the row/column structure. This means that only the first 768 bytes of each of the first 768 rows will be changed in arr1.

我写了一个for循环，但它不够快满足我的需求。

I wrote a for loop for this, but it's not fast enough for my needs.

for (int x = 0; x < 768; x++) //copy each row
{
    memcpy(arr1+(1024*x),arr2+(768*x), nc);
}

有没有更好的解决办法？

Is there a better solution?

推荐答案

也许摆脱了乘法

size_t bigindex = 0, smallindex = 0;
for (int x = 0; x < 768; x++) //copy each row
{
    memcpy(arr1 + bigindex, arr2 + smallindex, nc);
    bigindex += 1024;
    smallindex += 768;
}

---

修改 D'哦！使用指针！

unsigned char *a1 = arr1;
unsigned char *a2 = arr2;
for (int x = 0; x < 768; x++) //copy each row
{
    memcpy(a1, a2, nc);
    a1 += 1024;
    a2 += 768;
}

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