3

exp1/matmult.c

@@ -0,0 +1,163 @@
+/*
+任务5(选做）：编写一个N×N矩阵乘法函数的并行线程化版本，程序保存为matmult.c，
+设计一种方案，验证并行程序正确性。
+编译、调试、运行程序，设2^k ≤ N < 2^(k+1)
+给出线程数为1、2、4、…、2^(k-1)、2^k等情况下的运行时间，计算加速比与效率，并对结果给出解释。
+*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <pthread.h>
+#include <sys/time.h>
+
+double get_time()
+{
+    struct timeval tv;
+    gettimeofday(&tv, NULL);
+    return tv.tv_sec + tv.tv_usec / 1000000.0;
+}
+
+typedef struct {
+    int id;
+    int start_row;
+    int end_row;
+    int N;
+    double *A;
+    double *B;
+    double *C;
+} thread_arg_t;
+
+void *mul_thread(void *arg)
+{
+    thread_arg_t *ta = (thread_arg_t *)arg;
+    for (int i = ta->start_row; i < ta->end_row; i++)
+    {
+        for (int j = 0; j < ta->N; j++)
+        {
+            double sum = 0.0;
+            for (int k = 0; k < ta->N; k++)
+            {
+                sum += ta->A[i * ta->N + k] * ta->B[k * ta->N + j];
+            }
+            ta->C[i * ta->N + j] = sum;
+        }
+    }
+    return NULL;
+}
+
+/* 串行版本用于验证 */
+void matmul_serial(double *A, double *B, double *C, int N)
+{
+    for (int i = 0; i < N; i++)
+    {
+        for (int j = 0; j < N; j++)
+        {
+            double sum = 0.0;
+            for (int k = 0; k < N; k++)
+            {
+                sum += A[i * N + k] * B[k * N + j];
+            }
+            C[i * N + j] = sum;
+        }
+    }
+}
+
+int verify(double *C1, double *C2, int N)
+{
+    for (int i = 0; i < N * N; i++)
+    {
+        if (C1[i] - C2[i] > 1e-6 || C2[i] - C1[i] > 1e-6)
+        {
+            printf("Mismatch at index %d: %f vs %f\n", i, C1[i], C2[i]);
+            return 0;
+        }
+    }
+    return 1;
+}
+
+int main(int argc, char **argv)
+{
+    int N, nthreads;
+
+    if (argc != 3)
+    {
+        printf("Usage: %s <N> <nthreads>\n", argv[0]);
+        exit(0);
+    }
+    N = atoi(argv[1]);
+    nthreads = atoi(argv[2]);
+
+    double *A = (double *)malloc(N * N * sizeof(double));
+    double *B = (double *)malloc(N * N * sizeof(double));
+    double *C = (double *)malloc(N * N * sizeof(double));
+    double *C_serial = (double *)malloc(N * N * sizeof(double));
+
+    srand(42);
+    for (int i = 0; i < N * N; i++)
+    {
+        A[i] = (double)(rand() % 100) / 10.0;
+        B[i] = (double)(rand() % 100) / 10.0;
+    }
+
+    /* 并行版本计时 */
+    pthread_t *tid = (pthread_t *)malloc(nthreads * sizeof(pthread_t));
+    thread_arg_t *targs = (thread_arg_t *)malloc(nthreads * sizeof(thread_arg_t));
+
+    double t_start = get_time();
+    int rows_per_thread = N / nthreads;
+    int extra = N % nthreads;
+    int current_row = 0;
+
+    for (int i = 0; i < nthreads; i++)
+    {
+        targs[i].id = i;
+        targs[i].start_row = current_row;
+        targs[i].end_row = current_row + rows_per_thread + (i < extra ? 1 : 0);
+        targs[i].N = N;
+        targs[i].A = A;
+        targs[i].B = B;
+        targs[i].C = C;
+        current_row = targs[i].end_row;
+        pthread_create(&tid[i], NULL, mul_thread, &targs[i]);
+    }
+
+    for (int i = 0; i < nthreads; i++)
+    {
+        pthread_join(tid[i], NULL);
+    }
+    double t_end = get_time();
+    double t_parallel = t_end - t_start;
+
+    /* 串行版本验证 */
+    double t_ser_start = get_time();
+    matmul_serial(A, B, C_serial, N);
+    double t_ser_end = get_time();
+    double t_serial = t_ser_end - t_ser_start;
+
+    printf("Matrix size: %d x %d, Threads: %d\n", N, N, nthreads);
+    printf("Serial time:   %f seconds\n", t_serial);
+    printf("Parallel time: %f seconds\n", t_parallel);
+
+    if (verify(C, C_serial, N))
+    {
+        printf("Verification: PASSED - parallel result matches serial.\n");
+    }
+    else
+    {
+        printf("Verification: FAILED.\n");
+    }
+
+    if (t_parallel > 0)
+    {
+        printf("Speedup: %f\n", t_serial / t_parallel);
+        printf("Efficiency: %f\n", t_serial / (nthreads * t_parallel));
+    }
+
+    free(A);
+    free(B);
+    free(C);
+    free(C_serial);
+    free(tid);
+    free(targs);
+    return 0;
+}