int*d_out,intN){intthreads=128;intblocks=min((N+threads-1)/threads,MAX_BLOCKS);device_copy_scalar_kernel<<<blocks,threads>>>(d_in,d_out,N);}
这个kernel是Triton的fused softmax 教程改过来的,在那个教程里 Triton 的速度比 PyTorch 和 torch.compile 都要快,所以这里的性能表现似乎有点奇怪,因为两者都是element-wise操作。接着作者把上面的BLOCK_SIZE固定为1024,观察到性能有很大提升 这里如果固定了BLOCK_SIZE,那上面的Kernel也要做对应的修改比如以BLOCK_SI...
size_t size = min( int(prop.l2CacheSize * 0.75) , prop.persistingL2CacheMaxSize ); cudaDeviceSetLimit( cudaLimitPersistingL2CacheSize, size); // set-aside 3/4 of L2 cache for persisting accesses or the max allowed size_t window_size = min(prop.accessPolicyMaxWindowSize, num_bytes); ...
In that case, the valid range for (width, height, depth) is ((1,maxTexture2DGather[0]), (1,maxTexture2DGather[1]), 0). CUDA array typeValid extents that must always be met {(width range in elements), (height range), (depth range)} Valid extents with cudaArraySurfaceLoadStore ...
deviceProp.maxGridSize[0], deviceProp.maxGridSize[1], deviceProp.maxGridSize[2]); 如果一个SM中存在多个线程束调度器,则同时一个SM中会有多个线程束被调度。 通常一个GPU并行计算的完整简单流程为: 在主机端申请适当大小的显存用于存放输入数据以及接收输出,将主机端的数据拷贝到设备端; ...
[i];}// in only one thread, process final element (if there is one)if(idx==N/2&&N%2==1)d_out[N-1]=d_in[N-1];}voiddevice_copy_vector2(int*d_in,int*d_out,intn){threads=128;blocks=min((N/2+threads-1)/threads,MAX_BLOCKS);device_copy_vector2_kernel<<<blocks,threads>>...
samplesreal matrix of shape [number of samples, number of features] or list of real matrices which are rbind()-ed internally. In the latter case, is is possible to pass in more than INT32_MAX samples. centroidsreal matrix with precalculated clusters' centroids (e.g., using kmeans() or...
(void*)d_z, nBytes, cudaMemcpyDeviceToHost); // 检查执行结果 float maxError = 0.0; for (int i = 0; i < N; i++) maxError = fmax(maxError, fabs(z[i] - 30.0)); std::cout << "最大误差: " << maxError << std::endl; // 释放device内存 cudaFree(d_x); cudaFree(d_y)...
Description Calculates half max(a, b) defined as (a > b) ? a : b. ‣ If either of inputs is NaN, the other input is returned. ‣ If both inputs are NaNs, then canonical NaN is returned. ‣ If values of both inputs are 0.0, then +0.0 > -0.0 __device__ __half __...
其中⊗可表示为求 sum,min,max,avg 等操作。 对于数组归约的并行计算问题,要从一个数组出发,最终得到一个数。假如数组元素个数是 2 的整数次方,可以将数组后半部分的各个元素与前半部分对应的数组元素相加,在GPU的并行计算中采用tree-based计算方法经过n轮迭代将数据做累加,最后得到sum。这就是所谓的折半归约...