56fd68e7dd
CCCL (CUDA C++ Core Libraries) provides: - CUB: device/block/warp-level GPU primitives (reduce, scan, sort, topk) - Thrust: high-level parallel algorithms (transform_reduce, sort, scan) - libcudacxx: CUDA C++ standard library (atomics, barriers, memory) - cudax: experimental features (memory resources, allocators) - Tuning policies: per-SM hardware-specific algorithm parameters Competition optimization vectors mapped to CCCL: - Output TPS (83% weight): warp_reduce, block_reduce, device_topk - Input TPS (14% weight): device_scan, block_load, prefetch - Cache TPS (3% weight): prefix caching strategy patterns - Memory (0.9 util): pooled/cached/buddy allocators Source: https://github.com/NVIDIA/cccl (shallow clone, HEAD only) License: Apache-2.0
53 lines
1.5 KiB
Plaintext
53 lines
1.5 KiB
Plaintext
//===----------------------------------------------------------------------===//
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//
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// Part of CUDASTF in CUDA C++ Core Libraries,
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// under the Apache License v2.0 with LLVM Exceptions.
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// See https://llvm.org/LICENSE.txt for license information.
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// SPDX-License-Identifier: Apache-2.0 WITH LLVM-exception
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// SPDX-FileCopyrightText: Copyright (c) 2022-2024 NVIDIA CORPORATION & AFFILIATES.
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//
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//===----------------------------------------------------------------------===//
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/**
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* @file
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*
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* @brief Approximate pi using Monte Carlo method
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*
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*/
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#include <cuda/experimental/stf.cuh>
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#include <curand_kernel.h>
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#include <stdio.h>
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using namespace cuda::experimental::stf;
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int main(int, char**)
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{
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context ctx;
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auto lsum = ctx.logical_data(shape_of<scalar_view<size_t>>());
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size_t N = 1000000;
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ctx.parallel_for(box(N), lsum.reduce(reducer::sum<size_t>{}))->*[] __device__(size_t i, auto& sum) {
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curandState local_state;
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curand_init(1234, i, 0, &local_state);
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double x = curand_uniform_double(&local_state); // Random x in [0, 1)
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double y = curand_uniform_double(&local_state); // Random y in [0, 1)
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// Count (x,y) coordinates which are within the unit circle
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if (x * x + y * y <= 1.0)
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{
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sum++;
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}
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};
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// We get the ratio of "shots" within the unit circle and the total number of
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// "shots". The surface of the quarter of unit circle [0, 1) x [0, 1) is pi/4
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auto res = ctx.wait(lsum);
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double pi_val = (4.0 * res) / N;
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ctx.finalize();
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_CCCL_ASSERT(fabs(pi_val - 3.1415) < 0.1, "Invalid result");
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}
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