2

我在处理 CUDA 设备上的 2D 数组时遇到了一个奇怪的问题。

    #define VR 100 // rows
    #define ST 13 // columns
    __global__ void test(float *arr, curandState *globalState, size_t pitch, unsigned long seed) {
    int id = (blockIdx.x * blockDim.x)  + threadIdx.x;
    curand_init ( seed, id, 0, &globalState[id] );
    cuPrintf("Thread id: %d \n", id);

    float* row = (float*)(((char*)arr) + id * pitch);
    for (int j = 0; j < ST; ++j) {
        row[j] = generate(globalState, id);
    }

}

int main() {
    float *d_arr;
    float *h_arr = new float[VR*ST];
    size_t pitch;
    cudaMallocPitch(&d_arr, &pitch, ST* sizeof(float), VR);

    dim3 dimBlock(VR); 
    dim3 dimGrid(1,1);

    curandState* devStates;
    cudaMalloc ( &devStates, VR*ST*sizeof( curandState ) );

    test <<< dimGrid, dimBlock >>> (d_arr, devStates, pitch, unsigned(time(NULL)));
    cudaMemcpy(h_arr, d_arr,VR*ST*sizeof(float),cudaMemcpyDeviceToHost);

    for (int i=0; i<VR; i++) {
        for (int j=0; j<ST; j++) {
            cout << "N["<<i<<"]["<<j<<"]=" << h_arr[(i*ST)+j]<<endl;
        }
    }

我没有得到均匀分布的数字，而是它们以 13 的顺序出现，中间有一堆零。见： http: //pastie.org/6106381

4

2 回答 2

4

问题是原始数据数组是使用分配的，cudaMallocPitch而复制是使用普通的cudaMemcpy。这将产生意想不到的结果，因为该cudaMallocPitch操作会创建“填充”行以满足对齐要求，而 cudaMemcpy 假定所有内容都是连续存储的。以下是我认为具有更正功能的代码：

    #include <stdio.h>
    #include <iostream>
    #include <curand_kernel.h>

    #define VR 100 // rows
    #define ST 13 // columns


__device__ float generate(curandState* globalState, int id)
{
    //int id = (blockIdx.x * blockDim.x)  + threadIdx.x;
    curandState localState = globalState[id];
    float rand;
    do {
        rand = curand_uniform( &localState );
    } while(rand == 0); //
    globalState[id] = localState;
    return rand;
}


    __global__ void test(float *arr, curandState *globalState, size_t pitch, unsigned long seed) {
    int id = (blockIdx.x * blockDim.x)  + threadIdx.x;
    curand_init ( seed, id, 0, &globalState[id] );
    //cuPrintf("Thread id: %d \n", id);

    float* row = (float*)(((char*)arr) + id * pitch);
    for (int j = 0; j < ST; ++j) {
        row[j] = generate(globalState, id);
    }

}

    using namespace std;
int main() {
    float *d_arr;
    float *h_arr = new float[VR*ST];
    size_t pitch;
    cudaMallocPitch(&d_arr, &pitch, ST* sizeof(float), VR);

    dim3 dimBlock(VR);
    dim3 dimGrid(1,1);

    curandState* devStates;
    cudaMalloc ( &devStates, VR*ST*sizeof( curandState ) );

    test <<< dimGrid, dimBlock >>> (d_arr, devStates, pitch, unsigned(time(NULL)));
    cudaMemcpy2D(h_arr, ST*sizeof(float),  d_arr, pitch, ST*sizeof(float), VR ,cudaMemcpyDeviceToHost);

    for (int i=0; i<VR; i++) {
        for (int j=0; j<ST; j++) {
            cout << "N["<<i<<"]["<<j<<"]=" << h_arr[(i*ST)+j]<<endl;
        }
    }
}

使用以下代码编译上述代码：

nvcc -arch=sm_20 -lcurand  -o t70 t70.cu

然后运行我得到看似“正常”的输出：

N[0][0]=0.876772
N[0][1]=0.550017
N[0][2]=0.49023
N[0][3]=0.530145
N[0][4]=0.501616
N[0][5]=0.326232
N[0][6]=0.438308
N[0][7]=0.857651
N[0][8]=0.462743
N[0][9]=0.38252
N[0][10]=0.258212
N[0][11]=0.194021
N[0][12]=0.895522
N[1][0]=0.559201
N[1][1]=0.257747
N[1][2]=0.430971
N[1][3]=0.707209
N[1][4]=0.599081
N[1][5]=0.0457626
N[1][6]=0.702412
N[1][7]=0.88791
N[1][8]=0.508877
N[1][9]=0.702734
N[1][10]=0.379898
N[1][11]=0.138841
N[1][12]=0.540869

（结果被截断）

于 2013-02-10T00:30:45.590 回答
0

我认为这是错误的，您应该分配 VR 线程数或块数，因为您已经在内核中循环了 ST。

也许这会解决它。

于 2013-02-09T22:22:38.753 回答