我正在尝试使用VS2013在VB.NET中使用CUDAFy 1.29。
我试图从CUDAFy翻译C#的样本,我有两种类型的错误,如下所述:
我的变量
Shared cs_CC As String = "adiciona"
Shared MyGPU As GPGPU = Nothing
Shared Arch As eArchitecture = Nothing
我的代码:
Shared Executa()
if Loader = true
Dim Modulo = CudafyModule.TryDeserialize(cs_cc)
If IsNothing(Modulo) OrElse (Not Modulo.TryVerifyChecksums) Then
Modulo = CudafyTranslator.Cudafy(ePlatform.All, Arch, cs_CC.GetType)
Modulo.Serialize()
End If
MyGPU.Loadmodule(Modulo)
Dim a As Integer() = New Integer(N - 1) {}
Dim b As Integer() = New Integer(N - 1) {}
Dim c As Integer() = New Integer(N - 1) {}
' allocate the memory on the GPU
Dim dev_a As Integer() = MyGPU.Allocate(Of Integer)(a)
Dim dev_b As Integer() = MyGPU.Allocate(Of Integer)(b)
Dim dev_c As Integer() = MyGPU.Allocate(Of Integer)(c)
' fill the arrays 'a' and 'b' on the CPU
For i As Integer = 0 To N - 1
a(i) = i
b(i) = 2 * i
Next
' copy the arrays 'a' and 'b' to the GPU
MyGPU.CopyToDevice(a, dev_a)
MyGPU.CopyToDevice(b, dev_b)
For i As Integer = 0 To 128
MyGPU.Launch(1, 1).adiciona(dev_a, dev_b, dev_c)
Next
end if
End Sub
将在CUDA上运行的ADICIONA函数
<Cudafy()> _
Shared Sub adiciona(thread As GThread, a As Integer(), b As Integer(), c As Integer())
Dim tid As Integer = thread.blockIdx.x
While tid < N
c(tid) = a(tid) + b(tid)
tid += thread.gridDim.x
End While
End Sub
LOADER:尝试识别卡和CUDA(成功运行):
Public Shared Function Loader() As Boolean
DeviceType = eGPUType.Cuda
CudafyModes.Target = DeviceType
CudafyTranslator.Language = If(CudafyModes.Target = eGPUType.Cuda, eLanguage.Cuda, eLanguage.OpenCL)
Dim CompatibleDevice As GPGPUProperties() = CudafyHost.GetDeviceProperties(CudafyModes.Target, True).ToArray
If Not CompatibleDevice.Any Then ' não possui um full-CUDA device
MsgBox("I do not found any OpenCL or CUDA compatible device")
Return False
End If
Dim selectedDevice As GPGPUProperties = CompatibleDevice(0)
If IsNothing(selectedDevice) Then
MsgBox("I cannot allocate a compatible device")
Return False
End If
CudafyModes.DeviceId = selectedDevice.DeviceId
Thread_per_Block = selectedDevice.MaxThreadsPerBlock
Blocks_per_Grid = selectedDevice.MaxThreadsSize.x
Shared_Mem_per_Block = selectedDevice.SharedMemoryPerBlock
MyGPU = CudafyHost.GetDevice(CudafyModes.Target, CudafyModes.DeviceId)
Arch = MyGPU.GetArchitecture
Return True
End Function
问题:
问题1:如果我只使用 dim Modulo作为CudafyModule = CudafyTranslator.Cudafy(),我会收到以下错误:不支持检查语句。< / strong>很奇怪! Cudafy的所有文档都完全按照它显示这一行!
问题2:因此,我尝试检查是否存在已写入的模块(.CDFY),如果不存在,则调用Serialize()函数。问题是该函数在我的文件夹中创建一个名为 STRING.CDFY 的文件,但不是 ADICIONA.CDFY ,也不是 ADD_LONG_LOOP.CDFY 。是对的。由于我想避免在每次运行此代码时编译,如何正确地使CUDAFy编写它?
问题3:当VS运行时,一切正常,直到调用ADICIONA(MyGPU.Launch)! VS通过消息“在模块中找不到'ADICIONA'”来停止执行。
有趣的是注意到:
1-创建两个临时文件(.CU,.PTX)以及.CDFY文件。它证明了NVCC编译器运行良好并创建了CUDA模块。那么,为什么代码没有找到函数ADICIONA?
2-用C#编写的所有样本都在这里运行100%。从C#和VB的转换似乎没问题(我用过TELERIK来做)。我不认为问题可能与此有关,但我可能错了。
3-问题与NVCC编译器和VB.NET的一些引用无关,因为代码编译。
我试着用CODEPLEX写一个答案。根本没有答案...... 我试图在Internet上看到很多样本,但所有这些都是为CUDAFy C#创建的,并且没有一个使用1.29和CUDA 7.5版。
另外,我想了解为什么基本功能(CudafyTranslator.Cudafy())在VB中出错但在C#中没有。
那么,有没有人使用VB.NET成功创建了CUDAFy代码?
非常感谢您的帮助。
答案 0 :(得分:0)
要解决问题1,请选中项目属性&gt;下的“删除整数溢出检查”。编译&gt;高级编译选项&gt;优化。更多信息:http://forums.asp.net/post/715302.aspx
答案 1 :(得分:0)
尝试以下简单示例:
Imports Cudafy
Imports Cudafy.Host
Imports Cudafy.Translator
Public Class Form1
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim GPU As GPGPU = CudafyHost.GetDevice(eGPUType.OpenCL, CudafyModes.DeviceId)
Dim Modulo As CudafyModule = CudafyTranslator.Cudafy(ePlatform.All, GPU.GetArchitecture())
GPU.LoadModule(Modulo)
Dim Resultado As Integer
Dim ResultadoGPU As Integer() = gpu.Allocate(Of Integer)()
GPU.Launch().Adicionar(2, 7, ResultadoGPU)
GPU.CopyFromDevice(ResultadoGPU, Resultado)
MessageBox.Show("2 + 7 = " & Resultado)
GPU.Launch().Subtrair(2, 7, ResultadoGPU)
GPU.CopyFromDevice(ResultadoGPU, Resultado)
MessageBox.Show("2 - 7 = " & Resultado)
GPU.Free(ResultadoGPU)
End Sub
<Cudafy()>
Private Shared Sub Adicionar(a As Integer, b As Integer, c As Integer())
c(0) = a + b
End Sub
<Cudafy()>
Private Shared Sub Subtrair(a As Integer, b As Integer, c As Integer())
c(0) = a - b
End Sub
End Class