110 lines
4.3 KiB
Plaintext
110 lines
4.3 KiB
Plaintext
#include <iostream>
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#include <omp.h>
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#include <chrono>
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#include <vector>
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#include <iomanip>
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#include <cmath>
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void matrixMultiplyCPU(const float* A, const float* B, float* C, int M, int N, int K, int num_threads) {
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#pragma omp parallel for num_threads(num_threads)
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for (int i = 0; i < M; ++i) {
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for (int j = 0; j < K; ++j) {
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float sum = 0.0f;
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for (int k = 0; k < N; ++k) {
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sum += A[i * N + k] * B[k * K + j];
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}
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C[i * K + j] = sum;
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}
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}
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}
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void runCPUTest() {
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std::vector<int> matrix_sizes = {256, 512, 1024, 2048};
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std::vector<int> thread_counts = {8, 64, 256};
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std::cout << "CPU矩阵乘法性能测试 (OpenMP多线程)\n";
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std::cout << "=================================================================\n";
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std::cout << std::setw(12) << "Matrix"
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<< std::setw(12) << "Threads"
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<< std::setw(15) << "Time(ms)"
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<< std::setw(15) << "FLOPS(G)"
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<< std::setw(15) << "Speedup" << std::endl;
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std::cout << "-----------------------------------------------------------------\n";
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// 存储基准性能（单线程）
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std::vector<double> baseline_times(matrix_sizes.size());
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for (size_t m = 0; m < matrix_sizes.size(); ++m) {
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int size = matrix_sizes[m];
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int M = size, N = size, K = size;
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// 分配内存
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float *A = new float[M * N];
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float *B = new float[N * K];
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float *C = new float[M * K];
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// 初始化数据
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for (int i = 0; i < M * N; ++i) A[i] = (rand() % 100) / 100.0f;
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for (int i = 0; i < N * K; ++i) B[i] = (rand() % 100) / 100.0f;
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// 首先测试单线程作为基准
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auto start = std::chrono::high_resolution_clock::now();
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matrixMultiplyCPU(A, B, C, M, N, K, 1);
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auto end = std::chrono::high_resolution_clock::now();
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auto single_duration = std::chrono::duration<float, std::milli>(end - start).count();
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baseline_times[m] = single_duration;
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// 测试多线程
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for (int threads : thread_counts) {
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start = std::chrono::high_resolution_clock::now();
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matrixMultiplyCPU(A, B, C, M, N, K, threads);
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end = std::chrono::high_resolution_clock::now();
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auto duration = std::chrono::duration<float, std::milli>(end - start).count();
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// 计算FLOPS
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double total_flops = 2.0 * M * N * K;
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double gflops = total_flops / (duration * 1e6);
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// 计算加速比
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double speedup = baseline_times[m] / duration;
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std::cout << std::setw(12) << size << "x" << size
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<< std::setw(12) << threads
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<< std::setw(15) << std::fixed << std::setprecision(3) << duration
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<< std::setw(15) << std::fixed << std::setprecision(2) << gflops
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<< std::setw(15) << std::fixed << std::setprecision(2) << speedup << std::endl;
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}
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delete[] A;
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delete[] B;
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delete[] C;
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std::cout << "-----------------------------------------------------------------\n";
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}
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}
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void plotData() {
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std::cout << "\n\nASCII图表：CPU性能分析\n";
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std::cout << "=================================================================\n";
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std::cout << "1. 不同线程数下的加速比趋势\n";
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std::cout << " Matrix Threads=8 Threads=64 Threads=256\n";
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// 这里可以添加具体的绘图逻辑
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// 由于是文本输出，可以使用简单的ASCII字符绘制柱状图
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std::cout << "\n2. 不同矩阵规模下的性能趋势\n";
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std::cout << " Threads 256x256 512x512 1024x1024 2048x2048\n";
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std::cout << "\n注意：完整图表建议使用Python (matplotlib) 生成。\n";
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std::cout << "推荐生成以下图表：\n";
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std::cout << "- 折线图：不同线程数下的加速比 vs 矩阵规模\n";
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std::cout << "- 柱状图：不同配置下的GFLOPS对比\n";
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std::cout << "- 热力图：线程数 × 矩阵规模 的性能分布\n";
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}
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int main() {
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runCPUTest();
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plotData();
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return 0;
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}
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