Question

我开始使用cuda，我正在尝试一个简单的例子，我将两个数组发送到一个全局函数，一个复制到另一个，然后返回第二个。

我有：

__global__
void add(int n, int *tri, int *y)
{
    int index = threadIdx.x;
    int stride = blockDim.x;
    for (int i = index; i < n; i += stride)
        y[i] = tri[i];
}

和

   //local copy of data
    int *tri2 = tri; // data checked, and is valid

    int *y = new int[width * height]; // same size as `tri`
    int N = width * height;

    // Allocate Unified Memory – accessible from CPU or GPU
    cudaMallocManaged(&tri2, N * sizeof(int));
    cudaMallocManaged(&y, N * sizeof(int));

    // initialize y array on the host
    for (int i = 0; i < N; i++) {
        y[i] = 2;
    }

    // Run kernel on the GPU
    add << <1, 256 >> >(N, tri2, y);

    // Wait for GPU to finish before accessing on host
    cudaDeviceSynchronize();

    //copy back to host
    int i = 0;
    int f = -999.0; /* CPU copy of value */
    cudaMemcpy(&f, &y[i], sizeof(int), cudaMemcpyDeviceToHost);

    std::cout << "back: " << f << std::endl;
    std::cout << "orig: " << tri[i] << std::endl;

orig值为128，与进入时相同。返回的f值始终为0.我缺少什么？

Answer 1

数组tri的值与数组tri2的值不同。

使用

cudaMallocManaged(&tri2, N * sizeof(int));

你在设备上分配新内存，我假设恰好是零。然后在内核中将这个零数组复制到y。永远不会复制数组tri的值。

以下是一些如何做到这一点的示例。（另）

int* tri = ....

int* tri_managed;
//allocate new managed memory and save pointer in tri_managed
cudaMallocManaged(&tri_managed, sizeof(int) * N);
//now copy entries of tri to tri_managed
for(int i = 0; i < N; i++)
    tri_managed[i] = tri[i];

int* y;
cudaMallocManaged(&y, N * sizeof(int));

// initialize y array
for (int i = 0; i < N; i++) {
    y[i] = 2;
}

//copy entries of tri_managed to y
add << <1, 256 >> >(N, tri_managed, y);

// Wait for GPU to finish before accessing on host
cudaDeviceSynchronize();

//copy back to host
int i = 0;
int f = -999.0; /* CPU copy of value */
//cudaMemcpy(&f, &y[i], sizeof(int), cudaMemcpyDeviceToHost);
//since managed memory is accessible on host and device, we can just to this
f = y[i];

std::cout << "back: " << f << std::endl;
std::cout << "orig: " << tri[i] << std::endl;

//don't forget to free memory after usage
cudaFree(tri_managed);
cudaFree(y);

从全局cuda函数返回数据？

1 个答案: