Cuda blockDim.y 总是 ==1

Question

我总是得到 blockdim.y ==1。无论我在 numBlocks 中设置什么值，我总是得到相同的值。

__global__ void CalcVideo(unsigned char *original, unsigned char *candidate, int *answer)
{
    printf("block id.x = %d blockid.y=%d blockdim.x = %d blockdim.y = %d Thread id= %d \n", 
        blockIdx.x, blockIdx.y, blockDim.x, blockDim.y, threadIdx.x );
}

int ORIGINAL_FRAMES = 3;
int CANDIDATE_FRAMES = 2;
int FRAME_LENGHT = 3;

dim3 numBlocks(ORIGINAL_FRAMES, CANDIDATE_FRAMES);
    dim3 threadsPerBlock(3);  // 64 threads

CalcVideo << <numBlocks, threadsPerBlock >> >(original_device, candidate_device, answer_device);

Num of y.blokcs 执行正确，但为什么程序给我错误的 blockdim.y 大小？

block id.x = 1 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 0
block id.x = 1 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 1
block id.x = 1 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 2
block id.x = 1 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 0
block id.x = 1 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 1
block id.x = 1 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 2
block id.x = 0 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 0
block id.x = 0 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 1
block id.x = 0 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 2
block id.x = 0 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 0
block id.x = 0 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 1
block id.x = 0 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 2
block id.x = 2 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 0
block id.x = 2 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 1
block id.x = 2 blockid.y=1 blockdim.x = 3 blockdim.y = 1 Thread id= 2
block id.x = 2 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 0
block id.x = 2 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 1
block id.x = 2 blockid.y=0 blockdim.x = 3 blockdim.y = 1 Thread id= 2

Answer 1

blockDim 存储一个块的维度。在您的情况下，您将 threadsPerBlock 作为块维度传递，这使其成为 3 x 1 x 1。内核调用的第一个参数，在你的例子中是 numBlocks，它控制块的网格的维度——你可以在内核中作为 gridDim 访问它。

旁注：我假设问题中极低的块数和大小仅用于测试目的，因为它们会使任何 GPU 在实践中极度未得到充分利用。

Cuda blockDim.y 总是 ==1

Cuda blockDim.y always ==1

c++

cuda

gpu