Question

我在CUDA中使用了全局2d数组变量，并且我尝试对此全局变量使用累积加法。但是当我重新运行代码时，它从上次运行的值开始。例如，如果上次运行中的值为50，则下一次运行将显示100.它不会将值重置为0.

__device__ double *d_t; 
__device__ size_t d_gridPitch;

__global__ void kernelFunc()
{
    int i = blockIdx.x * blockDim.x + threadIdx.x
    double* rowt = (double*)((char *)d_t + i * d_gridPitch);
    rowt[0] = rowt[0] + 50000;

    printf("%.0f, ",rowt[0]);
}


int main()
{
    int size = 16;
    size_t d_pitchLoc;
    double *d_tLoc;

    cudaMallocPitch((void**)&d_tLoc, &d_pitchLoc, size * sizeof(double), size);
    cudaMemcpyToSymbol(d_gridPitch, &d_pitchLoc, sizeof(int));
    cudaMemcpyToSymbol(d_t, & d_tLoc, sizeof(d_tLoc));

    kernelFunc<<<1,size>>>();
    cudaDeviceReset();

    return 0;
}

Answer 1

听起来你想要的是初始化你正在分配的内存（请注意，这与＃34;重置变量＆＃34;）无关。为此，请使用cudaMemset2D初始化cudaMallocPitch返回的内存分配中的字节。所以你的主机API序列如下所示：

int size = 16;
size_t d_pitchLoc;
double *d_tLoc;

cudaMallocPitch((void**)&d_tLoc, &d_pitchLoc, size * sizeof(double), size);
cudaMemset2D(d_tLoc, d_pitchLoc, 0, size * sizeof(double), size);

cudaMemcpyToSymbol(d_gridPitch, &d_pitchLoc, sizeof(int));
cudaMemcpyToSymbol(d_t, & d_tLoc, sizeof(d_tLoc));

请注意cudaMemset2D，如cudaMemset，sets bytes来自提供的int值的LSB。

如何重置CUDA全局变量

1 个答案: