Question

我试图将以下（简化的）嵌套循环移植为CUDA 2D内核。 NgS和NgO的大小会随着数据集的增加而增加;现在我只想让这个内核输出所有值的正确结果：

// macro that translates 2D [i][j] array indices to 1D flattened array indices
#define idx(i,j,lda) ( (j) + ((i)*(lda)) )

int NgS  = 1859;
int NgO  = 900;

// 1D flattened matrices have been initialized as:
 Radio_cpu = new double [NgS*NgO];
Result_cpu = new double [NgS*NgO];
// ignoring the part where they are filled w/ data

for (m=0; m<NgO; m++) {        
    for (n=0; n<NgS; n++) {
            Result_cpu[idx(n,m,NgO)]] = k0*Radio_cpu[idx(n,m,NgO)]];
    }
}

我遇到的例子通常处理方形循环，而且与CPU版本相比，我无法获得所有GPU数组索引的正确输出。这是调用内核的主机代码：

dim3 dimBlock(16, 16);
dim3 dimGrid;
dimGrid.x = (NgO + dimBlock.x - 1) / dimBlock.x;
dimGrid.y = (NgS + dimBlock.y - 1) / dimBlock.y;

// Result_gpu and Radio_gpu are allocated versions of the CPU variables on GPU
trans<<<dimGrid,dimBlock>>>(NgO, NgS, k0, Radio_gpu, Result_gpu);

这是内核：

__global__ void trans(int NgO, int NgS,
                      double k0, double * Radio, double * Result) {

int n = blockIdx.x * blockDim.x + threadIdx.x;
int m = blockIdx.y * blockDim.y + threadIdx.y;

if(n > NgS || m > NgO) return;

// map the two 2D indices to a single linear, 1D index
int grid_width = gridDim.x * blockDim.x;
int idxxx = m + (n * grid_width);

Result[idxxx] = k0 * Radio[idxxx];
}

使用当前代码，我在复制后继续将Result_cpu变量与Result_gpu变量进行比较。当我循环显示我得到的值时：

    // matches from NgS = 0...913
    Result_gpu[NgS = 913][NgO = 0]: -56887.2
    Result_cpu[Ngs = 913][NgO = 0]: -56887.2

    // mismatches from NgS = 914...1858
    Result_gpu[NgS = 914][NgO = 0]: -12.2352
    Result_cpu[NgS = 914][NgO = 0]: 79448.6

无论NgO的值如何，此模式都是相同的。我一直试图通过查看几个小时的各种示例并尝试更改来弄清楚我在哪里犯了错误，但到目前为止，这个方案已经减去了明显的问题，而其他方案已经导致内核调用错误/左GPU阵列未初始化为所有值。由于我显然无法看清错误，所以如果有人能指出我正确的方向，我真的很感激。我很确定它在我的鼻子底下，我无法看到它。

如果重要，我会在Kepler卡上测试此代码，使用MSVC 2010，CUDA 4.2和304.79驱动程序进行编译，并使用arch=compute_20,code=sm_20和arch=compute_30,code=compute_30标志编译代码没有区别。

Answer 1

@vaca_loca：我测试了以下内核（对我来说也适用于非方块尺寸）：

__global__ void trans(int NgO, int NgS,
                  double k0, double * Radio, double * Result) {

int n = blockIdx.x * blockDim.x + threadIdx.x;
int m = blockIdx.y * blockDim.y + threadIdx.y;
if(n > NgO || m > NgS) return;
int ofs = m * NgO + n;
Result[ofs] = k0 * Radio[ofs];
}

void test() {

int NgS  = 1859, NgO  = 900;
int data_sz = NgS * NgO, bytes = data_sz * sizeof(double);
cudaSetDevice(0);
double *Radio_cpu = new double [data_sz*3],
    *Result_cpu = Radio_cpu + data_sz,
    *Result_gpu = Result_cpu + data_sz;
double k0 = -1.7961233;

srand48(time(NULL));
int i, j, n, m;
for(m=0; m<NgO; m++) {
  for (n=0; n<NgS; n++) {
        Radio_cpu[m + n*NgO] = lrand48() % 234234;
        Result_cpu[m + n*NgO] = k0*Radio_cpu[m + n*NgO];
    }
}

double *g_Radio, *g_Result;
cudaMalloc((void **)&g_Radio, bytes * 2);
g_Result = g_Radio + data_sz;
cudaMemcpy(g_Radio, Radio_cpu, bytes, cudaMemcpyHostToDevice);

dim3 dimBlock(16, 16);
dim3 dimGrid;
dimGrid.x = (NgO + dimBlock.x - 1) / dimBlock.x;
dimGrid.y = (NgS + dimBlock.y - 1) / dimBlock.y;

trans<<<dimGrid,dimBlock>>>(NgO, NgS, k0, g_Radio, g_Result);

cudaMemcpy(Result_gpu, g_Result, bytes, cudaMemcpyDeviceToHost);

for(m=0; m<NgO; m++) {
    for (n=0; n<NgS; n++) {
        double c1 = Result_cpu[m + n*NgO],
                c2 = Result_gpu[m + n*NgO];
        if(std::abs(c1-c2) > 1e-4)
            printf("(%d;%d): %.7f %.7f\n", n, m, c1, c2);
    }
}
cudaFree(g_Radio);
delete []Radio_cpu;
}

但是，在我看来，使用四边形从全局内存访问数据可能不是非常缓存，因为访问步幅非常大。您可以考虑使用2D纹理，如果它对您的算法在2D位置访问数据至关重要

二维内核调用和非方矩阵的启动参数

1 个答案: