如何复制内存 [英] How to copy memory

问题描述

说我有

unsigned char *varA, *varB, *varC;
varA=malloc(64);
varB=malloc(32);
varC=malloc(32);

如何将第一个 32字节的varA放入varB，然后将最后一个 32字节的varA放入varC?

How can i put the first 32 byte of varA into varB and the last 32 byte of varA into varC?

推荐答案

memcpy(varB, varA, 32);
memcpy(varC, varA + 32, 32);

就这么简单，因为基础数据类型是unsigned char，它的大小与字节相同.如果varA，varB和varC是整数，则需要将memcpy(即32)的size参数乘以sizeof(int)，以计算要复制的正确字节数.如果我在做书呆子，我可以在上面的示例中将sizeof(unsigned char)乘以32，但这不是必需的，因为sizeof(unsigned char) == 1.

It's this simple because the underlying data type is unsigned char, which is the same size as a byte. If varA, varB, and varC were integers, you would need to multiply the size parameter to memcpy (i.e. 32) by sizeof(int) to compute the right number of bytes to copy. If I were being pedantic, i could have multiplied 32 by sizeof(unsigned char) in the example above, but it is not necessary because sizeof(unsigned char) == 1.

请注意，我不需要将varA + 32中的32乘以任何值，因为当向指针添加常量偏移量时，编译器会为我执行此操作.

Note that I don't need to multiply the 32 in varA + 32 by anything because the compiler does that for me when adding constant offsets to pointers.

另一件事:如果想提高速度，只需要单独处理varA的每一半就足够了，而不是分配两个新的缓冲区并复制到其中.

One more thing: if you want to be fast, it might be sufficient to just work on each half of varA separately, rather than allocate two new buffers and copy into them.

这篇关于如何复制内存的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！