使用线程查找给定范围的质数 [英] Using Threads To Find Prime Numbers of a Given Range

问题描述

请帮助我

我正在尝试执行以下操作:

创建一个程序来计算从某个数字X到一个数字Y的素数.修改该程序，使其同时运行两个线程.第一个线程将从X到Y开始计数，第二个线程从Y到Z开始计数. "

我设法创建了程序并使线程运行.我唯一的问题是，对于大于90000的输入值，该程序将不再可靠，例如对于100000-150000的范围，程序可能只输出1个质数，如果再次运行，则输出5个质数，而另一个时间输出3个质数.
我已经在没有线程的情况下测试了该程序，并且运行良好，这使我认为这是我对错误线程的实现.

如果可以的话，请帮助我.谢谢

到目前为止，这是我的代码:

Please help me out

I am trying to do the following:

"Create a program that calculates prime numbers from a certain number X to a number Y. Modify the program so as to run two threads at the same time. The first thread will start counting from X to Y and the second from Y to Z."

I have managed to create the program and get the threads running. My only problem is that for input values greater than 90000, the program is no longer reliable e.g. for a range of 100000 - 150000, the program might output only 1 prime number and if run again output 5 prime numbers and yet another time output 3 prime numbers.
I have tested the program without threads and it works fine, which leads me to think that it''s my implementation of the thread that is faulty.

Please help me out if you can. Thanks

here is my code so far:

#include <iostream>
#include <stdio.h>
#include <windows.h>

using namespace std;
const int MAX = 10000;

struct prArray
{
    int pStart, pEnd, pTest, pCount;
    int tArray[MAX];
};

int check_prime(int);
DWORD WINAPI primer(LPVOID lpParam);

int main()
{
    /* by having these 2 variables + the struct, error checkin is made easier*/
    int one, two, diff, temp, end1, start2;

    cout << " START:\n";
    cin >> one;
    cin >> two;
    prArray ash;
    prArray jeff;
    diff = (two - one); // get the difference between the start and end values
    temp = diff / 2;        // use this to divide the work evenly among the threads
    end1 = one + temp;      // end value for first thread
    start2 = end1 + 1;      // begin 1 number after the end of first thread

    // set the struct values
    ash.pStart = one;       // equal to the first number of the range e.g. 1 - 10000 --> 1
    ash.pEnd = one + temp;  // equal to halfway between the range e.g. 5000
    ash.pCount = ash.pStart;
    jeff.pStart = start2;   // equal to next number after halfway e.g. 5001
    jeff.pEnd = two;        // equal to last number of the range e.g. 10000
    jeff.pCount = jeff.pStart;


    //create first thread
    DWORD myThreadID1;
    HANDLE myHandle = CreateThread(
        NULL,           // security
        0,              // stack size
        primer,         // function to perform
        &ash,           // function parameter --> pass the struct
        0,              // creation flag
        &myThreadID1);  // thread id

    //create second thread
    DWORD myThreadID2;
    HANDLE myHandle2 = CreateThread(
        0,
        0,
        primer,
        &jeff,
        0,
        &myThreadID2);

    // close the threads
    CloseHandle(myHandle);
    CloseHandle(myHandle2);


    // print stuff out
    cout << endl << endl;
    cout << "   BACK TO main()\n";
    // begin with first struct (ash)
    cout << "\n   THREAD1: \n";
    cout << "Thread1 starting point: " << ash.pStart << endl;
    cout << "Thread1 ending point: " << ash.pEnd << endl;
    cout << "Test: " << ash.pTest << endl;
    int i1;
    for (i1 = 0; i1 < MAX; i1++) // print array values for first struct (ash)
    {
        if (ash.tArray[i1] == 0)
        {
            break;
        }
        else
        {
            cout << ash.tArray[i1] << " ";
        }
    }

    //begin second thread (jeff)
    cout << endl << endl;
    cout << "   THREAD2: " << endl;
    cout << "Thread2 starting point: " << jeff.pStart << endl;
    cout << "Thread2 ending point: " << jeff.pEnd << endl;
    cout << "Test: " << jeff.pTest << endl;
    int js;
    for (js = 0; js < MAX; js++) // print array values for second struct (jeff)
    {
        if (jeff.tArray[js] == 0)
        {
            break;
        }
        else
        {
            cout << jeff.tArray[js] << " ";
        }
    }
    cout << "\n first = " << i1 << endl;
    cout << " second = " << js << endl;
    cout << "Thread 1 counter = " << ash.pCount << endl;
    cout << "Thread 2 counter = " << jeff.pCount << endl;
    cout << "\n END PROGRAM" << endl;
    system("pause");
    return 0;
}
// =============    END OF main() ==============

/** ============    CHECK FOR PRIME ============= */
int check_prime(int a)
{
    int c;
    if ( a == 2 || a == 3 || a == 5 || a ==7) //check if the number is 2
    {
        return 1;
    }
    else if (a % 2 == 0 || a % 3 == 0 || a % 5 == 0 || a % 7 == 0) //check if the number is even
    {
        return 0;
    }
    else //for odd numbers, check if they are prime
    {
        for (c = 11; c <= a - 1; c+=2)
        {
            if (a%c == 0)
            return 0;
        }
        if ( c == a)
            return 1;
    }
}
/** ============    END CHECK PRIME============== */

/** ============    THREAD  ===================== */
DWORD WINAPI primer(LPVOID lpParam)
{
    prArray& dan = *(prArray*)lpParam; /*working initialization */
    dan.pTest = (dan.pEnd - dan.pStart);

    // initialize the array to "0"
    for (int i = 0; i < MAX; i++)
    {
        dan.tArray[i] = 0;
    }

    int b = 0;
    for (int lp = dan.pStart; lp <= dan.pEnd; lp++, dan.pCount++)
    {
        int result3 = check_prime(lp);
        if (result3 == 1)
        {
            dan.tArray[b] = lp;
            b++;
        }
    }
}
/** ================END THREAD =================*/

推荐答案

您不必等待线程结束. 您可以使用以下几行来等待2个线程的结束:

You do not wait the end of your threads.
You might use the following lines to wait the end of your 2 threads :

HANDLE handles[2];

handles[0]=myHandle;
handles[1]=myHandle2;
WaitForMultipleObjects(2,handles,TRUE,INFINITE);

CloseHandle(myHandle);
CloseHandle(myHandle2);

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