减少C Amp中的GPU-CPU数据传输

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.

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)