将字节数组转换为向量的最有效方法

问题描述

将16个字节的数组转换为最有效的方法是什么uint4向量?目前，我手动将字节或整理为uint，然后进行设置具有完整uint的向量分量.是否有OpenCL支持执行此任务?

What is the most efficient way of converting an array of 16 bytes intoa uint4 vector ? currently, I manually OR the bytes into uints, then setthe vector's components with the completed uints. Is there OpenCL support for performing this task?

这是针对OpenCL 1.2

This is for OpenCL 1.2

这是我的代码:

    local uchar buffer[16];
    uint v[4];
    for (int i = 0; i < 4; ++i) {
        v[i]=0;
        for (int j = 0; j < 4; ++j) {
            v[i] |= (buffer[(i<<2)+j]) << (j<<3);
        }
    }
    uint4 result = (uint4)(v[0],v[1],v[2],v[3]);

缓冲区实际上是本地缓冲区.

Edit 2: buffer is actually a local buffer.

推荐答案

您应该能够即时进行转换，而无需复制数据:

You should be able to convert it on the fly without copying the data:

local uchar buffer[16];
if(get_local_id(0) == 0)
{
    for (int x = 0; x < 4; ++x)
    {
        buffer[x] = x + 1;
        buffer[x + 4] = x + 2;
        buffer[x + 8] = x + 3;
        buffer[x + 12] = x + 4;
    }
    local uint4 *result = (local uint4*)buffer;
    printf("0x%x 0x%x 0x%x 0x%x\n", (*result).x, (*result).y, (*result).z, (*result).w);
}

结果:

0x4030201 0x5040302 0x6050403 0x7060504

如果您仍然需要复制数据，请执行以下操作:

If you need to copy the data though you do:

uint4 result = *(local uint4*)buffer;

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