尝试使用C ++ Amp优化我的应用程序时遇到以下问题:数据传输。对我来说,将数据从CPU复制到GPU没有问题(因为我可以在应用程序的初始状态下执行此操作)。更糟糕的是,我需要快速访问由C ++ Amp内核计算的结果,因此GPU和CPU之间的瓶颈是一个痛苦。我读到Windows 8.1下的性能提升,但是我使用的是Windows 7,我并不打算改变它。我读到了关于暂存数组但我不知道他们如何帮助解决我的问题。我需要向主机返回一个浮点值,这似乎是最耗时的操作。
float Subset::reduction_cascade(unsigned element_count, concurrency::array<float, 1>& a)
{
static_assert(_tile_count > 0, "Tile count must be positive!");
//static_assert(IS_POWER_OF_2(_tile_size), "Tile size must be a positive integer power of two!");
assert(source.size() <= UINT_MAX);
//unsigned element_count = static_cast<unsigned>(source.size());
assert(element_count != 0); // Cannot reduce an empty sequence.
unsigned stride = _tile_size * _tile_count * 2;
// Reduce tail elements.
float tail_sum = 0.f;
unsigned tail_length = element_count % stride;
// Using arrays as a temporary memory.
//concurrency::array<float, 1> a(element_count, source.begin());
concurrency::array<float, 1> a_partial_result(_tile_count);
concurrency::parallel_for_each(concurrency::extent<1>(_tile_count * _tile_size).tile<_tile_size>(), [=, &a, &a_partial_result] (concurrency::tiled_index<_tile_size> tidx) restrict(amp)
{
// Use tile_static as a scratchpad memory.
tile_static float tile_data[_tile_size];
unsigned local_idx = tidx.local[0];
// Reduce data strides of twice the tile size into tile_static memory.
unsigned input_idx = (tidx.tile[0] * 2 * _tile_size) + local_idx;
tile_data[local_idx] = 0;
do
{
tile_data[local_idx] += a[input_idx] + a[input_idx + _tile_size];
input_idx += stride;
} while (input_idx < element_count);
tidx.barrier.wait();
// Reduce to the tile result using multiple threads.
for (unsigned stride = _tile_size / 2; stride > 0; stride /= 2)
{
if (local_idx < stride)
{
tile_data[local_idx] += tile_data[local_idx + stride];
}
tidx.barrier.wait();
}
// Store the tile result in the global memory.
if (local_idx == 0)
{
a_partial_result[tidx.tile[0]] = tile_data[0];
}
});
// Reduce results from all tiles on the CPU.
std::vector<float> v_partial_result(_tile_count);
copy(a_partial_result, v_partial_result.begin());
return std::accumulate(v_partial_result.begin(), v_partial_result.end(), tail_sum);
}
我检查过上面的例子中最耗时的操作是copy(a_partial_result, v_partial_result.begin());
。我想找到一个更好的方法。
答案 0 :(得分:1)
所以我认为这里还有其他事情发生。您是否尝试过运行代码所基于的原始示例?这是available on CodePlex。
加载样本解决方案并在Release模式下构建Reduction项目,然后在没有附加调试器的情况下运行它。你应该看到这样的输出。
Running kernels with 16777216 elements, 65536 KB of data ...
Tile size: 512
Tile count: 128
Using device : NVIDIA GeForce GTX 570
Total : Calc
SUCCESS: Overhead 0.03 : 0.00 (ms)
SUCCESS: CPU sequential 9.48 : 9.45 (ms)
SUCCESS: CPU parallel 5.92 : 5.89 (ms)
SUCCESS: C++ AMP simple model 25.34 : 3.19 (ms)
SUCCESS: C++ AMP simple model using array_view 62.09 : 20.61 (ms)
SUCCESS: C++ AMP simple model optimized 25.24 : 1.81 (ms)
SUCCESS: C++ AMP tiled model 29.70 : 7.27 (ms)
SUCCESS: C++ AMP tiled model & shared memory 30.40 : 7.56 (ms)
SUCCESS: C++ AMP tiled model & minimized divergence 25.21 : 5.77 (ms)
SUCCESS: C++ AMP tiled model & no bank conflicts 25.52 : 3.92 (ms)
SUCCESS: C++ AMP tiled model & reduced stalled threads 21.25 : 2.03 (ms)
SUCCESS: C++ AMP tiled model & unrolling 22.94 : 1.55 (ms)
SUCCESS: C++ AMP cascading reduction 20.17 : 0.92 (ms)
SUCCESS: C++ AMP cascading reduction & unrolling 24.01 : 1.20 (ms)
请注意,没有任何示例在您编码的时间附近。虽然可以说CPU更快,数据复制时间是这里的一个重要因素。
这是可以预料的。有效使用GPU涉及将诸如还原之类的操作转移到GPU。您需要移动大量计算以弥补复制开销。
你应该考虑的一些事情:
更多有用的信息