c# - clrobj() 将结构数组传递给 GPU 内核（ALEA 库）时没有 llvm

Question

对于我尝试使用 ALEA 库将结构数组传递给 NVIDIA 内核的代码，我收到“Fody/Alea.CUDA：clrobj(cGPU) 没有 llvm”构建错误。这是我的代码的简化版本。我删除了输出收集功能以保持代码简单。我现在只需要能够将结构数组发送到 GPU。

using Alea.CUDA;
using Alea.CUDA.Utilities;
using Alea.CUDA.IL;

namespace GPUProgramming
{
  public class cGPU
  {
   public int Slice;
   public float FloatValue;
  }

  [AOTCompile(AOTOnly = true)]
  public class TestModule : ILGPUModule
  {
    public TestModule(GPUModuleTarget target) : base(target)
    {
    }

    const int blockSize = 64;

    [Kernel]
    public void Kernel2(deviceptr<cGPU> Data, int n)
    {
      var start = blockIdx.x * blockDim.x + threadIdx.x;
      int ind = threadIdx.x;

      var sharedSlice =         __shared__.Array<int>(64);
      var sharedFloatValue =    __shared__.Array<float>(64);

      if (ind < n && start < n)
      {
        sharedSlice[ind] = Data[start].Slice;
        sharedFloatValue[ind] = Data[start].FloatValue;
        Intrinsic.__syncthreads();
      }
    }

    public void Test2(deviceptr<cGPU> Data, int n, int NumOfBlocks)
    {
      var GridDim = new dim3(NumOfBlocks, 1);
      var BlockDim = new dim3(64, 1);

      try
      {
        var lp = new LaunchParam(GridDim, BlockDim);
        GPULaunch(Kernel2, lp, Data, n);
      }
      catch (CUDAInterop.CUDAException x)
      {
        var code = x.Data0;
        Console.WriteLine("ErrorCode = {0}", code);
      }
    }
    public void Test2(cGPU[] Data)
    {
      int NumOfBlocks = Common.divup(Data.Length, blockSize);
      using (var d_Slice = GPUWorker.Malloc(Data))
      {
        try
        {
          Test_Kernel2(d_Slice.Ptr, Data.Length, NumOfBlocks);
        }
        catch (CUDAInterop.CUDAException x)
        {
          var code = x.Data0;
          Console.WriteLine("ErrorCode = {0}", x.Data0);
        }
      }
    }
  }
}

Answer 1

您的数据是类，即引用类型。尝试使用结构。引用类型不太适合 Gpu，因为它需要在堆上分配小内存。