C++ 中的简单时间比较多线程程序(和 4 个问题)

时间:2021-07-07 19:57:41

标签: c++ multithreading visual-c++ thread-safety

我有 C++ 方面的经验,我正在尝试使用该语言学习多线程。 我刚刚写了下面的程序(问题下面的代码)来比较 10 个函数调用和并行运行的时间效率。

我的四个问题是:

  1. 这是线程库的正确用法吗?时间似乎合理,因为线程应该快得多,但我是这个功能的新手,想确定我做对了。如果程序有任何改进,请告诉我。

  2. 是预期的输出(在代码下方)吗?线程试图同时打印到控制台,因此在新行之前写入了一些字符或使用其他值打印的字符,例如:3 + 5 = 84 + 5 = (新行) 9,而不是of 3 + 5 = 8 (new line) 4 + 5 = 9。这种行为是预期的吗?

  3. 关于这个主题的任何类型的阅读材料或建议也将不胜感激!我一直在阅读文章,并计划很快观看一些有关多线程的视频。

  4. 每次运行输出时间都不一样。当然,这是意料之中的,但有时,多线程运行比逐个函数调用运行要慢。这是应该发生的吗?


#include <iostream>
#include <thread>
#include <time.h>
#include <chrono>
#include <ctime>

using namespace std;

void add(int num) {
    cout << num << " + 5 = " << num + 5 << endl;
}

int main()
{
    int a, b, c, d, e, f, g, h, i, j;
    a = 0;
    b = 1;
    c = 2;
    d = 3;
    e = 4;
    f = 5;
    g = 6;
    h = 7;
    i = 8;
    j = 9;
    
    cout << "Starting timer for no multithreading" << endl;

    std::chrono::high_resolution_clock::time_point t1 = std::chrono::high_resolution_clock::now();
    add(a);
    add(b);
    add(c);
    add(d);
    add(e);
    add(f);
    add(g);
    add(h);
    add(i);
    add(j);
    std::chrono::high_resolution_clock::time_point t2 = std::chrono::high_resolution_clock::now();

    cout << "Stopped timer for no multithreading" << endl;

    std::chrono::duration<double> total1 = std::chrono::duration_cast<std::chrono::duration<double>>(t2 - t1);

    cout << "Without multithreading, the ten function calls took: " << total1.count() << " seconds to complete." << endl;
    

    cout << endl << endl;

    thread A(add, a);
    thread B(add, b);
    thread C(add, c);
    thread D(add, d);
    thread E(add, e);
    thread F(add, f);
    thread G(add, g);
    thread H(add, h);
    thread I(add, i);
    thread J(add, j);

    cout << "Starting timer for multithreading" << endl;

    std::chrono::high_resolution_clock::time_point t3 = std::chrono::high_resolution_clock::now();
    A.join();
    B.join();
    C.join();
    D.join();
    E.join();
    F.join();
    G.join();
    H.join();
    I.join();
    J.join();
    std::chrono::high_resolution_clock::time_point t4 = std::chrono::high_resolution_clock::now();

    cout << "Stopped timer for multithreading" << endl;

    std::chrono::duration<double> total2 = std::chrono::duration_cast<std::chrono::duration<double>>(t4 - t3);

    cout << "With multithreading, the ten function calls took: " << total2.count() << " seconds to complete." << endl;
    //cout << total2 << "seconds" << endl;
    

    return 0;
}

输出是:

Starting timer for no multithreading
0 + 5 = 5
1 + 5 = 6
2 + 5 = 7
3 + 5 = 8
4 + 5 = 9
5 + 5 = 10
6 + 5 = 11
7 + 5 = 12
8 + 5 = 13
9 + 5 = 14
Stopped timer for no multithreading
Without multithreading, the ten function calls took: 0.0016732 seconds to complete.


0 + 5 = 5
1 + 5 = 6
2 + 5 = 7
3 + 5 = 84 + 5 =
9
5 + 5 = 106 + 5 = 11
7 + 5 =
12
8 + 5 = 13
Starting timer for multithreading9 + 5 = 14

Stopped timer for multithreading
With multithreading, the ten function calls took: 5.07e-05 seconds to complete.

提前致谢:D

1 个答案:

答案 0 :(得分:0)

这不是测量时间的正确方法。

thread A(add, a);

在这里,将创建一个线程对象,它可能会立即执行该函数(取决于操作系统调度程序)。

A.join();

在这里,您正在等待线程完成。

您的基本流程是

thread A(add, a);
start timer
A.join()
end timer.

因此,您的计时器测量了等待线程完成 + 执行几个不走运的线程所花费的时间。

还有一点,std::cout 不是线程安全的。