我发现如果我的程序生成的值小于100’000,则单线程版本比多线程版本更快.显然我做错了什么.

我的代码如下：

#include <iostream>
#include <fstream>
#include <set>
#include <string>
#include <thread>
#include <mutex>
#include <shared_mutex>
 
using namespace std;
 
set<unsigned long long> primeContainer;
shared_mutex m;
 
void checkPrime(const unsigned long long p)
{
    if (p % 3 == 0)
        return;
 
    bool isPrime = true;
    for (set<unsigned long long>::const_iterator it = primeContainer.cbegin(); it != primeContainer.cend(); ++it)
    {
        if (p % *it == 0)
        {
            isPrime = false;
            break;
        }
        if (*it * *it > p) // check only up to square root
            break;
    }
 
    if (isPrime)
        primeContainer.insert(p);
}
 
void checkPrimeLock(const unsigned long long p)
{
    if (p % 3 == 0)
        return;
 
    bool isPrime = true;
    try
    {
        shared_lock<shared_mutex> l(m);
        for (set<unsigned long long>::const_iterator it = primeContainer.cbegin(); it != primeContainer.cend(); ++it)
        {
            if (p % *it == 0)
            {
                isPrime = false;
                break;
            }
            if (*it * *it > p)
                break;
        }
    }
    catch (exception& e)
    {
        cout << e.what() << endl;
        system("pause");
    }
 
    if (isPrime)
    {
        try
        {
            unique_lock<shared_mutex> l(m);
            primeContainer.insert(p);
        }
        catch (exception& e)
        {
            cout << e.what() << endl;
            system("pause");
        }
    }
}
 
void runLoopThread(const unsigned long long& l)
{
    for (unsigned long long i = 10; i < l; i += 10)
    {
        thread t1(checkPrimeLock,i + 1);
        thread t2(checkPrimeLock,i + 3);
        thread t3(checkPrimeLock,i + 7);
        thread t4(checkPrimeLock,i + 9);
        t1.join();
        t2.join();
        t3.join();
        t4.join();
    }
}
 
void runLoop(const unsigned long long& l)
{
    for (unsigned long long i = 10; i < l; i += 10)
    {
        checkPrime(i + 1);
        checkPrime(i + 3);
        checkPrime(i + 7);
        checkPrime(i + 9);
    }
}
 
void printPrimes(const unsigned long long& l)
{
    if (1U <= l)
        cout << "1 ";
    if (2U <= l)
        cout << "2 ";
    if (3U <= l)
        cout << "3 ";
    if (5U <= l)
        cout << "5 ";
 
    for (auto it = primeContainer.cbegin(); it != primeContainer.cend(); ++it)
    {
        if (*it <= l)
            cout << *it << " ";
    }
    cout << endl;
}
 
void writeToFile(const unsigned long long& l)
{
    string name = "primes_" + to_string(l) + ".txt";
    ofstream f(name);
 
    if (f.is_open())
    {
        if (1U <= l)
            f << "1 ";
        if (2U <= l)
            f << "2 ";
        if (3U <= l)
            f << "3 ";
        if (5U <= l)
            f << "5 ";
 
        for (auto it = primeContainer.cbegin(); it != primeContainer.cend(); ++it)
        {
            if (*it <= l)
                f << *it << " ";
        }
    }
    else
    {
        cout << "Error opening file." << endl;
        system("pause");
    }
}
 
int main()
{
    unsigned int n = thread::hardware_concurrency();
    std::cout << n << " concurrent threads are supported." << endl;
 
    unsigned long long limit;
    cout << "Please enter the limit of prime generation: ";
    cin >> limit;
 
    primeContainer.insert(7);
 
    if (10 < limit)
    {
        //runLoop(limit); //single-threaded
        runLoopThread(limit); //multi-threaded
    }
 
    printPrimes(limit);
    //writeToFile(limit);
    system("pause");
    return 0;
}

在main函数中,您将找到关于哪个函数是单线程和多线程的注释.

它们之间的主要区别在于锁的使用,容器迭代的共享以及插入的唯一性.如果重要,我的cpu有4个核心.

为什么单线程版本更快？