不使用memcpy复制字节 [英] Copying bytes without memcpy

问题描述

我在char数组中存储了几种不同类型的变量.通常我会这样写它们:

I have several variables of different types stored in a char array. Normally I would write them to the array this way:

int a = 5;
memcpy(offset, (char*)&a, sizeof(int))

但是，memcpy在OpenCL内核中不起作用.没有此功能，最简单的方法是什么?

However, memcpy doesn't work in OpenCL kernels. What would be the easiest way to do the same without this function?

推荐答案

您可以轻松地提供mymemcpy

You can easily enough provide mymemcpy

  void mymemcpy(unsigned char *dest, const unsigned char *src, size_t N)
  {
     size_t i;

     for(i=0;i<N;i++)
       dest[i] = src[i];
  }

但是效率不是很高，因为大多数副本都是4或8字节倍数的对齐副本.如果可以确定对齐方式为8个字节，则以unsigned long long为单位进行复制.有时甚至值得填充一个缓冲区以使其达到8个字节的倍数.

However it's not very efficient because most copies are aligned copies of multiples of 4 or 8 bytes. If you can work out that alignment is 8 bytes, copy in units of unsigned long long. Sometimes it's even worth padding a buffer to bring it up to a multiple of 8 bytes.

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