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编辑：在解决方案中编辑并编写了一些代码来帮助人们尝试实现嵌套的 for 循环。

所以我想在 cuda 内核中实现以下嵌套 for 循环：

for (unsigned in=1;in<=N;in++){         
     for (unsigned ie=1;ie<=N-1;ie+=2){
         // do some stuff with x[in and/or ie] and z[in and/or ie]
}

其中 x 和 z 是从主机复制的简单一维数组。

现在，就嵌套 for 循环实现而言，据我所知，这应该是相对简单的（如下面的 idx 和 idy 索引所示，有人可以确认吗？）。我的另一个问题是，如何在不使用任何类型的 if 语句和 % 的情况下将 ie 增加 2，而我的内核中我希望这会影响性能？

__global__ void myKernel(double x, double z, int N){

    // define indecies for storing counters
    int i = blockIdx.x * blockDim.x + threadIdx.x;
    int j = blockIdx.y * blockDim.y + threadIdx.y;

    // define counters
    int in = blockIdx.x * blockDim.x + (threadIdx.x + 1); // 1-based indexing
    int ie = (blockIdx.y * blockDim.y + (threadIdx.y)) * 2 + 1;

    z[i*N+j] = ie; // stores odd indicies [ie]

    // do some stuff with (in-ie), x[ie and/or in], z[ie and/or in]...
}

int main(){
    int N = 5; //arbitrary size of arrays
    int threadsPerBlock = 8; // semi-arbitrary, max 512 but depends on GPU device
    dim3 block(threadsPerBlock, threadsPerBlock/2 + 1); // launch N/2 + 1 threads in the second dimension since we are only interested in odd numbered indices.
    dim3 grid( (N-1+threadsPerBlock)/threadsPerBlock, (N-1+threadsPerBlock)/threadsPerBlock);

    std::vector<double> x;
    std::vector<double> z;
    z.resize(N*N);
    x.resize(N);

    //copy to device(d_x, d_z);

    //call the kernel
    myKernel<<<grid, block>>>(d_x, d_z, N);

    // retrieve from device

    return 0;
    }

谢谢

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1 回答 1

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代替：

int ie = blockIdx.y * blockDim.y + (threadIdx.y + 1);

你做这个 2n + 1：

int ie = (blockIdx.y * blockDim.y + threadIdx.y) * 2 + 1;

而且您显然只需要 N x N/2 个线程。

于 2013-09-09T08:16:18.690 回答