Question

所以我在我的代码中调用了2个函数的cudaMallocManaged，它在第一个函数（backwardMask（））之后工作正常我调用它后我可以轻松地从主机访问我的数据，但我的问题是内核函数seriesLength（） - 因为我在我的索引掩码上做cudaMallocManaged然后（在调用seriesLength（）之前）我可以轻松地在主机上访问/修改这个索引掩码，在我调用了seriesLength（）之后它正在调整我的索引掩码并且访问它也没有问题，但之后此函数返回我无法读取主机上的索引掩码并且异常（状态代码0xC0000022）。

这是一个非常奇怪的错误，因为我正在类似于第一个函数（backwardMask（）），它正常工作。

任何想法/解释都将受到高度赞赏。

这是seriesLengths内核函数代码：

__global__ void seriesLengths(int* scannedbw,int* indexmask,int* numOfSeries,int n){
int index = blockIdx.x * blockDim.x + threadIdx.x;
int stride = blockDim.x * gridDim.x;
 for (int i = index; i < n;i+=stride)
 {
    if (i == (n - 1))
    {
        *numOfSeries = scannedbw[i];
        indexmask[scannedbw[i]] = n;
    }
    if (i == 0)
    {
        indexmask[0] = 0;
    }
    else if (scannedbw[i] != scannedbw[i - 1])
    {
        indexmask[scannedbw[i] - 1] = i;
    }
 }
}

内核函数的函数backwardMask：

__global__ void backwardMask(const char *in, int* bwMask,int n)
{
    int index = blockIdx.x * blockDim.x + threadIdx.x;
    int stride = blockDim.x * gridDim.x;
    for (int i = index; i < n;i+=stride)
    {
        if (i == 0)
            bwMask[i] = 1;
        else 
        {
            bwMask[i] = (in[i] != in[i - 1]);
        }
    }
}

主要功能：

int main()
{
    int N=1024;
    srand(time(0));
    char* t;
    int* bwmask;
    cudaMallocManaged(&t, N*sizeof(char));
    cudaMallocManaged(&bwmask, N*sizeof(int));
    for (int i = 0; i < N; i++)
    {
        if(i<300)
        t[i] = 'a' + rand() % 2;
        else
            t[i] = 'a' + rand() % 20;

    }

    for (int j = 0; j < 60; j++)
        std::cout << t[j];
    std::cout << std::endl;
    int blockSize = 256;
    int numBlocks = (N + blockSize - 1) / blockSize;
    backwardMask<<<numBlocks, blockSize >>>(t,bwmask , N);
    cudaDeviceSynchronize();
    for (int j = 0; j < 60; j++)
        std::cout << bwmask[j];
    std::cout << std::endl;
    //now inclusive prefix sum for bwmask
    int* scannedbwmask;
    cudaMallocManaged(&scannedbwmask, N*sizeof(int));

    thrust::inclusive_scan(bwmask, bwmask + N, scannedbwmask);
    cudaDeviceSynchronize();

    int numOfSeries;
    //seriesLengths shows us lengths of each series by i-(i-1) and starting index of each series
    int* indexmask;
    cudaMallocManaged(&indexmask, (N+1)*sizeof(int));
    seriesLengths<<<numBlocks, blockSize>>>(scannedbwmask, indexmask, &numOfSeries, N);
    cudaDeviceSynchronize();

// accessing indexmask here gives us exception
    std::cout << indexmask[3];
    /*for (int j = 0; j < 60; j++)
        std::cout << indexmask[j];
    std::cout << std::endl;*/
    std::cout << "numseries " << numOfSeries;


    getch();
    return 0;
}

Answer 1

将numOfSeries更改为int

的指针

int* numOfSeries;

然后是malloc内存：

cudaMallocManaged(&numOfSeries, sizeof(int));

然后传递它：

seriesLengths<<<numBlocks, blockSize>>>(scannedbwmask, indexmask, numOfSeries, N);

使用它运行内核功能后无法访问统一内存

1 个答案: