如何解决C++ 中的简单时间比较多线程程序和 4 个问题
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:
-
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。
-
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?
#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
解决方法
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。
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