Question

程序利用线程计数到1000000000。没有互斥锁等，因为每个线程都访问相同的全局变量并且只是增加它。

// This is just a function pointer used later. Just for increasing the 
// readability on following code.
typedef void (*fptr)(int);

void* count_to(void*);
double stopwatch(fptr, int);

// Number to sum to
const long long MAX_SUM = 1000000000;
// Global sum that threads accesses
long long sum = 0;

// Main function creates threads from 1 to 11 and count the time
// in between executions using `stopwatch` function. 
int
main() {
    for (int i = 1; i < 12; ++i) {
        printf("\n======================================\n");
        printf("Time it took when thread count is %d : %f",
                i, stopwatch(create_and_execute, i));
        printf("\n======================================\n");
    }
    return EXIT_SUCCESS;
}

// Takes a function pointer and thread count
double
stopwatch(fptr func, int count) {
    clock_t begin = clock();
    func(count);
    clock_t end = clock();
    return (double)(end - begin) / CLOCKS_PER_SEC;
}

void
create_and_execute(count) {
    pthread_t* threads = (pthread_t*) malloc(sizeof(pthread_t) * count);

    if (threads == NULL) {
        fprintf(stderr, "Out of memory\n");
        exit(EXIT_FAILURE);
    }

    for (int i = 0; i < count; ++i) {
        if (pthread_create(&threads[i], NULL, count_to, NULL) != 0){
                fprintf(stderr, "Can't create thread %d\n", i);
                exit(EXIT_FAILURE);
        }
    }

    for (int i = 0; i < count; i++) {
        pthread_join(threads[i], NULL);
    }
    printf("Threads finished their job. Sum is: %lld\n", sum );
    sum = 0;

    free(threads);
}
// counts to MAX_SUM
void*
count_to(void* i) {
    while(sum < MAX_SUM) {
        sum += 1;
    }
    pthread_exit(0);
}

输出

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 1 : 2.044165
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 2 : 5.137091
======================================

======================================
Threads finished their job. Sum is: 1000000001
Time it took when thread count is 3 : 7.673170
======================================

======================================
Threads finished their job. Sum is: 1000000001
Time it took when thread count is 4 : 20.567974
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 5 : 25.316267
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 6 : 14.618212
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 7 : 34.998254
======================================

======================================
Threads finished their job. Sum is: 1000000002
Time it took when thread count is 8 : 40.366402
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 9 : 38.578370
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 10 : 30.201249
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 11 : 42.017907
======================================

问题在于，正如您所看到的，增加线程数会增加执行时间。除了一些例外。由于线程并行工作，我期望它会随着线程数的增加而大大减少执行时间。我确实理解，在一些线程计数之后增加线程数将保持与法律相同，正如人们在这里也可以观察到的那样。但为什么它增加而不是减少呢？难道我做错了什么？我完全错过了线程吗？任何建议都会有所帮助。

为了进一步说明我的问题，我的预期输出将是，例如：

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 1 : 2.044165
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 2 : 1.137091
======================================

======================================
Threads finished their job. Sum is: 1000000001
Time it took when thread count is 3 : 1.073170
======================================

======================================
Threads finished their job. Sum is: 1000000001
Time it took when thread count is 4 : 1.007974
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 5 : 0.316267
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 6 : 0.618212
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 7 : 0.998254
======================================

======================================
Threads finished their job. Sum is: 1000000002
Time it took when thread count is 8 : 0.366402
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 9 : 0.578370
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 10 : 0.201249
======================================

======================================
Threads finished their job. Sum is: 1000000000
Time it took when thread count is 11 : 0.017907
======================================

如果你认为某些事情不清楚，请告诉我！

PS：我最初在codereview中问过这个问题，但是有人建议它在这里更合适。所以，我在这里。

试图使用pthreads证明Amdahl定律的代码

0 个答案: