Question

我正在尝试进行FFT加内核计算。 FFT：managedCUDA库 kernel calc：自己的内核

C＃代码

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内核代码

public void cuFFTreconstruct() {
                CudaContext ctx = new CudaContext(0);
                CudaKernel cuKernel = ctx.LoadKernel("kernel_Array.ptx", "cu_ArrayInversion");

                float[] fData = new float[Resolution * Resolution * 2];
                float[] result = new float[Resolution * Resolution * 2];
                CudaDeviceVariable<float> devData = new CudaDeviceVariable<float>(Resolution * Resolution * 2);
                CudaDeviceVariable<float> copy_devData = new CudaDeviceVariable<float>(Resolution * Resolution * 2);

                int i, j;
                Random rnd = new Random();
                double avrg = 0.0;

                for (i = 0; i < Resolution; i++)
                {
                    for (j = 0; j < Resolution; j++)
                    {
                        fData[(i * Resolution + j) * 2] = i + j * 2;
                        fData[(i * Resolution + j) * 2 + 1] = 0.0f;
                    }
                }

                devData.CopyToDevice(fData);

                CudaFFTPlan1D plan1D = new CudaFFTPlan1D(Resolution * 2, cufftType.C2C, Resolution * 2);
                plan1D.Exec(devData.DevicePointer, TransformDirection.Forward);

                cuKernel.GridDimensions = new ManagedCuda.VectorTypes.dim3(Resolution / 256, Resolution, 1);
                cuKernel.BlockDimensions = new ManagedCuda.VectorTypes.dim3(256, 1, 1);

                cuKernel.Run(devData.DevicePointer, copy_devData.DevicePointer, Resolution);

                devData.CopyToHost(result);

                for (i = 0; i < Resolution; i++)
                {
                    for (j = 0; j < Resolution; j++)
                    {
                        ResultData[i, j, 0] = result[(i * Resolution + j) * 2];
                        ResultData[i, j, 1] = result[(i * Resolution + j) * 2 + 1];
                    }
                }   
                ctx.FreeMemory(devData.DevicePointer);
                ctx.FreeMemory(copy_devData.DevicePointer);
            }

但是这个程序效果不好。发生以下错误：

ErrorLaunchFailed：执行内核时设备发生异常。常见原因包括解除引用无效设备指针和访问超出范围的共享内存。不能使用上下文，因此必须销毁它（并且应该创建一个新的上下文）。此上下文中的所有现有设备内存分配均无效，如果程序要继续使用CUDA，则必须重新构建。

Answer 1

FFT计划将元素的数量，即复数的数量作为参数。因此，删除计划构造函数的第一个参数中的* 2。而批次数量的两倍也没有意义......

此外，我使用float2或cuFloatComplex类型（在ManagedCuda.VectorTypes中）来表示复数，而不是两个原始浮点数。要释放内存，请使用CudaDeviceVariable的Dispose方法。否则，GC稍后会在内部调用它。

主机代码看起来像这样：

int Resolution = 512;
CudaContext ctx = new CudaContext(0);
CudaKernel cuKernel = ctx.LoadKernel("kernel.ptx", "cu_ArrayInversion");

//float2 or cuFloatComplex
float2[] fData = new float2[Resolution * Resolution];
float2[] result = new float2[Resolution * Resolution];
CudaDeviceVariable<float2> devData = new CudaDeviceVariable<float2>(Resolution * Resolution);
CudaDeviceVariable<float2> copy_devData = new CudaDeviceVariable<float2>(Resolution * Resolution);

int i, j;
Random rnd = new Random();
double avrg = 0.0;

for (i = 0; i < Resolution; i++)
{
for (j = 0; j < Resolution; j++)
{
    fData[(i * Resolution + j)].x = i + j * 2;
    fData[(i * Resolution + j)].y = 0.0f;
}
}

devData.CopyToDevice(fData);

//Only Resolution times in X and Resolution batches
CudaFFTPlan1D plan1D = new CudaFFTPlan1D(Resolution, cufftType.C2C, Resolution);
plan1D.Exec(devData.DevicePointer, TransformDirection.Forward);

cuKernel.GridDimensions = new ManagedCuda.VectorTypes.dim3(Resolution / 256, Resolution, 1);
cuKernel.BlockDimensions = new ManagedCuda.VectorTypes.dim3(256, 1, 1);

cuKernel.Run(devData.DevicePointer, copy_devData.DevicePointer, Resolution);

devData.CopyToHost(result);

for (i = 0; i < Resolution; i++)
{
    for (j = 0; j < Resolution; j++)
    {
        //ResultData[i, j, 0] = result[(i * Resolution + j)].x;
        //ResultData[i, j, 1] = result[(i * Resolution + j)].y;
    }
}

//And better free memory using Dispose()
//ctx.FreeMemory is only meant for raw device pointers obtained from somewhere else...
devData.Dispose();
copy_devData.Dispose();
plan1D.Dispose();
//For Cuda Memory checker and profiler:
CudaContext.ProfilerStop();
ctx.Dispose();

Answer 2

感谢您提出此建议。

我尝试过建议的代码。但是，错误仍然存在。（错误：ErrorLaunchFailed：执行内核时设备发生异常。常见原因包括解除引用无效设备指针和访问超出范围的共享内存。上下文无法使用，因此必须销毁它（并且应该是新的）此上下文中的所有现有设备内存分配都是无效的，如果程序要继续使用CUDA，则必须重新构建。）

要使用float2，我按如下方式更改了cu代码

 extern "C"
{

__global__ void cu_ArrayInversion(float2* data_A, float2* data_B, int Resolution)
    {
    int image_x = blockIdx.x * blockDim.x + threadIdx.x;
    int image_y = blockIdx.y;

    data_B[(Resolution * image_x + image_y)].x = data_A[(Resolution * image_y + image_x)].x;
    data_B[(Resolution * image_x + image_y)].y = data_A[(Resolution * image_y + image_x)].y;
}

当程序执行“cuKernel.Run”时，进程停止。

ptx文件

.version 4.3
.target sm_20
.address_size 32

    // .globl   cu_ArrayInversion
.global .texref texref;

.visible .entry cu_ArrayInversion(
    .param .u32 cu_ArrayInversion_param_0,
    .param .u32 cu_ArrayInversion_param_1,
    .param .u32 cu_ArrayInversion_param_2
)
{
    .reg .f32   %f<5>;
    .reg .b32   %r<17>;


    ld.param.u32    %r1, [cu_ArrayInversion_param_0];
    ld.param.u32    %r2, [cu_ArrayInversion_param_1];
    ld.param.u32    %r3, [cu_ArrayInversion_param_2];
    cvta.to.global.u32  %r4, %r2;
    cvta.to.global.u32  %r5, %r1;
    mov.u32     %r6, %ctaid.x;
    mov.u32     %r7, %ntid.x;
    mov.u32     %r8, %tid.x;
    mad.lo.s32  %r9, %r7, %r6, %r8;
    mov.u32     %r10, %ctaid.y;
    mad.lo.s32  %r11, %r10, %r3, %r9;
    shl.b32     %r12, %r11, 3;
    add.s32     %r13, %r5, %r12;
    mad.lo.s32  %r14, %r9, %r3, %r10;
    shl.b32     %r15, %r14, 3;
    add.s32     %r16, %r4, %r15;
    ld.global.v2.f32    {%f1, %f2}, [%r13];
    st.global.v2.f32    [%r16], {%f1, %f2};
    ret;
}

Answer 3

感谢您的留言。

主机代码

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
using System.Drawing.Imaging;
using ManagedCuda;
using ManagedCuda.CudaFFT;
using ManagedCuda.VectorTypes;


namespace WFA_CUDA_FFT
{
    public partial class CuFFTMain : Form
    {
        float[, ,] FFTData2D;
        int Resolution;

        const int cuda_blockNum = 256;

        public CuFFTMain()
        {
            InitializeComponent();
            Resolution = 1024;
        }

        private void button1_Click(object sender, EventArgs e)
        {
            cuFFTreconstruct();
        }
        public void cuFFTreconstruct()
        {
            CudaContext ctx = new CudaContext(0);
            ManagedCuda.BasicTypes.CUmodule cumodule = ctx.LoadModule("kernel.ptx");
            CudaKernel cuKernel = new CudaKernel("cu_ArrayInversion", cumodule, ctx);
            float2[] fData = new float2[Resolution * Resolution];
            float2[] result = new float2[Resolution * Resolution];
            FFTData2D = new float[Resolution, Resolution, 2];
            CudaDeviceVariable<float2> devData = new CudaDeviceVariable<float2>(Resolution * Resolution);
            CudaDeviceVariable<float2> copy_devData = new CudaDeviceVariable<float2>(Resolution * Resolution);

            int i, j;
            Random rnd = new Random();
            double avrg = 0.0;

            for (i = 0; i < Resolution; i++)
            {
                for (j = 0; j < Resolution; j++)
                {
                    fData[i * Resolution + j].x = i + j * 2;
                    avrg += fData[i * Resolution + j].x;
                    fData[i * Resolution + j].y = 0.0f;
                }
            }

            avrg = avrg / (double)(Resolution * Resolution);

            for (i = 0; i < Resolution; i++)
            {
                for (j = 0; j < Resolution; j++)
                {
                    fData[(i * Resolution + j)].x = fData[(i * Resolution + j)].x - (float)avrg;
                }
            }

            devData.CopyToDevice(fData);

            CudaFFTPlan1D plan1D = new CudaFFTPlan1D(Resolution, cufftType.C2C, Resolution);
            plan1D.Exec(devData.DevicePointer, TransformDirection.Forward);

            cuKernel.GridDimensions = new ManagedCuda.VectorTypes.dim3(Resolution / cuda_blockNum, Resolution, 1);
            cuKernel.BlockDimensions = new ManagedCuda.VectorTypes.dim3(cuda_blockNum, 1, 1);

            cuKernel.Run(devData.DevicePointer, copy_devData.DevicePointer, Resolution);

            copy_devData.CopyToHost(result);

            for (i = 0; i < Resolution; i++)
            {
                for (j = 0; j < Resolution; j++)
                {
                    FFTData2D[i, j, 0] = result[i * Resolution + j].x;
                    FFTData2D[i, j, 1] = result[i * Resolution + j].y;
                }
            }

            //Clean up
            devData.Dispose();
            copy_devData.Dispose();
            plan1D.Dispose();
            CudaContext.ProfilerStop();
            ctx.Dispose();
        }
    }
}

内核代码

//Includes for IntelliSense 
#define _SIZE_T_DEFINED
#ifndef __CUDACC__
#define __CUDACC__
#endif
#ifndef __cplusplus
#define __cplusplus
#endif


#include <cuda.h>
#include <device_launch_parameters.h>
#include <texture_fetch_functions.h>
#include "float.h"
#include <builtin_types.h>
#include <vector_functions.h>
#include <vector>

// Texture reference
texture<float2, 2> texref;

extern "C"
{
    // Device code

    __global__ void cu_ArrayInversion(float2* data_A, float2* data_B, int Resolution)
    {
        int image_x = blockIdx.x * blockDim.x + threadIdx.x;
        int image_y = blockIdx.y;

        data_B[(Resolution * image_x + image_y)].y = data_A[(Resolution * image_y + image_x)].x;
        data_B[(Resolution * image_x + image_y)].x = data_A[(Resolution * image_y + image_x)].y;
    }
}

首先我用.Net4.5编译。此程序不起作用，并显示错误（System.BadImageFormatException）。但是当FFT函数注释掉时，内核程序会运行。

第二，我从.Net 4.5转向.Net 4.0。 FFT函数有效，但内核不运行并显示错误。

我的电脑是Windows 8.1专业版，我使用的是visual studio 2013。

使用managedCUDA进行1D FFT加内核计算

3 个答案: