CUDA在执行过程中组合线程独立（??）变量
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伙计们，如果标题令人困惑，我深表歉意。我虽然费时又辛苦，却无法想出正确的方式在一行中表达问题。因此，这里有更多详细信息。我正在做一个基本的图像减法，其中第二幅图像已被修改，我需要找到对该图像进行了多少更改的比率。为此，我使用了以下代码。两张图片均为128x1024。
for(int i = 0; i < 128; i++)
{
for(int j = 0; j < 1024; j++)
{
den++;
diff[i * 1024 + j] = orig[i * 1024 + j] - modified[i * 1024 + j];
if(diff[i * 1024 + j] < error)
{
num++;
}
}
}
ratio = num/den;
__global__ void calcRatio(float *orig, float *modified, int size, float *result)
{
int index = threadIdx.x + blockIdx.x * blockDim.x;
if(index < size)
result[index] = orig[index] - modified[index];
}
for(int i = 0; i < 128; i++)
{
for(int j = 0; j < 1024; j++)
{
if(modified[i * 1024 + j] < 400.0) //400.0 threshold value to ignore noise
{
den++;
diff[i * 1024 + j] = orig[i * 1024 + j] - modified[i * 1024 + j];
if(diff[i * 1024 + j] < error)
{
num++;
}
}
}
}
ratio = num/den;
没有找到相关结果
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疏腔傻小雹
template <int blocksize> __global__ void calcRatio(float *orig, float *modified, int size, float *result, int *count, const float error) { __shared__ volatile float buff[blocksize]; int index = threadIdx.x + blockIdx.x * blockDim.x; int stride = blockDim.x * gridDim.x; int count = 0; for(int i=index; i<n; i+=stride) { val = orig[index] - modified[index]; count += (val < error); result[index] = val; } buff[threadIdx.x] = count; __syncthreads(); // Parallel reduction in shared memory using 1 warp if (threadId.x < warpSize) { for(int i=threadIdx.x + warpSize; i<blocksize; i+= warpSize) { buff[threadIdx.x] += buff[i]; if (threadIdx.x < 16) buff[threadIdx.x] +=buff[threadIdx.x + 16]; if (threadIdx.x < 8) buff[threadIdx.x] +=buff[threadIdx.x + 8]; if (threadIdx.x < 4) buff[threadIdx.x] +=buff[threadIdx.x + 4]; if (threadIdx.x < 2) buff[threadIdx.x] +=buff[threadIdx.x + 2]; if (threadIdx.x == 0) count[blockIdx.x] = buff[0] + buff[1]; } }宦哨抹存胳