【编程题】多线程 （C++）

题目

来做题了

问题描述

有4个线程和1个公共的字符数组。线程1的功能就是向数组输出A，线程2的功能就是向字符输出B，线程3的功能就是向数组输出C，线程4的功能就是向数组输出D。要求按顺序向数组赋值ABCDABCDABCD，ABCD的个数由线程函数1的参数指定。[注：C语言选手可使用WINDOWS SDK库函数]

接口说明：

void init(); //初始化函数
void Release(); //资源释放函数
unsignedint__stdcall ThreadFun1(PVOID pM) ; //线程函数1，传入一个int类型的指针，用于初始化输出A次数，资源需要线程释放
unsignedint__stdcall ThreadFun2(PVOID pM) ;//线程函数2，无参数传入
unsignedint__stdcall ThreadFun3(PVOID pM) ;//线程函数3，无参数传入
Unsigned int __stdcall ThreadFunc4(PVOID pM);//线程函数4，无参数传入
char g_write[1032]; //线程1,2,3,4按顺序向该数组赋值。

输入

10

输出

ABCDABCDABCDABCDABCDABCDABCDABCDABCDABCD

思路

具体的函数说明可以在这个博客中找到 windows.h

利用信号量，把多线程当成单线程来用。。。。

顺便说一下markdown的图片调整大小

<img src="URL" width="200" height="200" /> 

代码

这里给出网上一个大神写的代码分享一下

#include <iostream>#include <cstring>#include <cstdlib>#include <windows.h>using namespace std;HANDLE hThread1=NULL;HANDLE hThread2=NULL;HANDLE hThread3=NULL;HANDLE hThread4=NULL;HANDLE hMutex=NULL;char  g_write[1032];int   g_writeCnt=0;  //已输出字符个数int   g_cnt=0;       //ABCD已输出次数unsigned long __stdcall ThreadFun1(PVOID pM)//线程函数1，传入一个int类型的指针[取值范围：1 – 250，测试用例保证]，用于初始化输出A次数，资源需要线程释放{    int N=*(int*)pM;    while(1)    {        WaitForSingleObject(hMutex,INFINITE);        if(g_cnt==N)  {             //cout<<endl<<"time enough!"<<endl;            exit(0);        }        g_write[g_writeCnt]='A';        cout<<g_write[g_writeCnt];        g_writeCnt++;        g_cnt++;        ReleaseMutex(hMutex);        Sleep(10);//防止while(1) CPU飙升        ResumeThread(hThread2);        SuspendThread(hThread1);    }    return 0;}unsigned long __stdcall ThreadFun2(PVOID pM)//线程函数2，无参数传入{    while(1)    {        WaitForSingleObject(hMutex,INFINITE);        g_write[g_writeCnt]='B';        cout<<g_write[g_writeCnt];        g_writeCnt++;        ReleaseMutex(hMutex);        Sleep(10);        ResumeThread(hThread3);        SuspendThread(hThread2);    }    return 0;}unsigned long __stdcall ThreadFun3(PVOID pM)//线程函数3，无参数传入{    while(1)    {        WaitForSingleObject(hMutex,INFINITE);        g_write[g_writeCnt]='C';        cout<<g_write[g_writeCnt];        g_writeCnt++;        ReleaseMutex(hMutex);        Sleep(10);        ResumeThread(hThread4);        SuspendThread(hThread3);    }    return 0;}unsigned long __stdcall ThreadFun4(PVOID pM)//线程函数4，无参数传入{    while(1)    {        WaitForSingleObject(hMutex,INFINITE);        g_write[g_writeCnt]='D';        cout<<g_write[g_writeCnt];        g_writeCnt++;        ReleaseMutex(hMutex);        Sleep(10);        ResumeThread(hThread1);        SuspendThread(hThread4);    }    return 0;}void init()  //初始化进程函数{    int N=0;//1-125    cin>>N;    N=N>125?125:N;    N=N<1?1:N;    hMutex=CreateMutex(NULL,false,NULL);    hThread2=CreateThread(NULL,0,ThreadFun2,NULL,CREATE_SUSPENDED,NULL);    hThread3=CreateThread(NULL,0,ThreadFun3,NULL,CREATE_SUSPENDED,NULL);    hThread4=CreateThread(NULL,0,ThreadFun4,NULL,CREATE_SUSPENDED,NULL);    hThread1=CreateThread(NULL,0,ThreadFun1,(void*)&N,0,NULL);    HANDLE hThreads[4]={hThread1,hThread2,hThread3,hThread4};    WaitForMultipleObjects(4,hThreads,TRUE,5000);}void Release()// 资源释放{    CloseHandle(hThread1);    CloseHandle(hThread2);    CloseHandle(hThread3);    CloseHandle(hThread4);    CloseHandle(hMutex);}int main(void){   // freopen("in.txt","r",stdin);//    init();    Release();    return 0;}

代码好多好复杂。。。。为什么不能简单一些，于是我上午学习了一下最新的C++支持的多线程特性。可以用thread函数，这样就方便多了。

下面根据C++11中的thread函数进行操作。
这里主要是采用信号量mutex来保证顺序性的。

#include <iostream>       // std::cout#include <thread>         // std::thread#include <mutex>          // std::mutexusing namespace std;volatile int counter(0); // non-atomic countermutex mtx1;           // locks access to countermutex mtx2;           // locks access to countermutex mtx3;           // locks access to countermutex mtx4;           // locks access to countervoid displayForThread1(){    while(1)    {        if(counter>0)        {            if (mtx1.try_lock())            {                cout<<"A";                mtx2.unlock();                counter--;            }        }        else        {            break;        }    }}void displayForThread2(){    while(1)    {        if(counter>0)        {            if (mtx2.try_lock())            {                cout<<"B";                mtx3.unlock();                counter--;            }        }        else        {            break;        }    }}void displayForThread3(){    while(1)    {        if(counter>0)        {            if (mtx3.try_lock())            {                cout<<"C";                mtx4.unlock();                counter--;            }        }        else        {            break;        }    }}void displayForThread4(){    while(1)    {        if(counter>0)        {            if (mtx4.try_lock())            {                cout<<"D";                mtx1.unlock();                counter--;            }        }        else        {            break;        }    }}int main (int argc, const char* argv[]){    thread threads[4];    int number;    cin>>number;    counter = number*4-1;//这里要减1 否则会多输出一个线程    mtx2.lock();    mtx3.lock();    mtx4.lock();    threads[0] = thread(displayForThread1);    threads[1] = thread(displayForThread2);    threads[2] = thread(displayForThread3);    threads[3] = thread(displayForThread4);    for (auto& th : threads) th.join();//所有线程 join    return 0;}