简易版的进程池

1.使用匿名管道实现 进程池

#pragma oncetypedef void(*task_t)();void task1()
{cout << "游戏 刷新日志" << endl;
}void task2()
{cout << "游戏 刷新野区" << endl;
}
void task3()
{cout << "游戏 检测软件是否更新，如果需要，就提示用户" << endl;
}
void task4()
{cout << "游戏 用户释放技能，更新用户的HP和MP"<<endl;
}
void Loadtask(vector<task_t> *tasks)
{tasks->push_back(task1);tasks->push_back(task2);tasks->push_back(task3);tasks->push_back(task4);
}//以上是头文件//以下是程序；
//类似于内存池，每次使用都不再向系统申请进程，而是提前储备进程；
#include<cstdlib>
#include<cassert>
#include<time.h>
#include<sys/wait.h>
#include<sys/stat.h>
#include<iostream>
#include<vector>
#include<string>
#include<unistd.h>using namespace std;
#include"main.h"
const int processnum = 5;//子进程数量=5
vector<task_t> tasks;  //任务列表，全局变量//先描述
class channel
{
public:channel(int cmdfd ,pid_t slaverid,const string &processname):_cmdfd(cmdfd),_slaverid(slaverid),_processname(processname){}
public:int _cmdfd;	//发送任务的文件描述符pid_t _slaverid; //子进程的pidstring _processname;//子进程的名字——方便我们打印日志；
};void slaver()
{//read(0);//直接从标准输入读while (true){int cmdcode = 0;int n = read(0,&cmdcode,sizeof(int));//如果父进程不给子进程发任务？就会阻塞等待；if (n == sizeof(int)){//执行cmdcode 对应的任务列表；cout << getpid() << ":cmdcode:" << cmdcode << endl;//查看父进程给子进程派发的任务码；if (cmdcode > 0 && cmdcode < tasks.size()) //满足条件运行函数；tasks[cmdcode]();}if (n == 0){break;}}
}
//函数传参，需要输入型，输出型，输入输出型参数；
//输入：const &： const引用
//输出：用指针
//输入输出：&引用
void InitProcessPool(vector<channel>* channels)
{//1.初始化for (int i = 0; i < processnum; i++){int pipefd[2];//临时空间；int n = pipe(pipefd);assert(!n);//判断是否成功；(void)n;pid_t id = fork();if (id == 0)//子进程读{close(pipefd[1]);dup2(pipefd[0], 0);//重定向，从标准输入中读close(pipefd[0]);slaver();//执行相对应的任务；cout << "process: " << getpid() << "quit" << endl;//打印子进程pid；退出进程；exit(0);}//父进程写入close(pipefd[0]);//添加channel字段；string name = "process-" + to_string(i);channels->push_back(channel(pipefd[1], id, name));}
}
void Debug(const vector<channel> &channels)
{//testfor (const auto& c : channels){cout << c._cmdfd << " " << c._slaverid << " " << c._processname << endl;}
}
void Menu()											//菜单
{cout << "1.刷新日志" << endl;cout << "2.刷新野怪" << endl;cout << "3.检测更新" << endl;cout << "4.更新状态" << endl;cout << "0.退出游戏" << endl;
}
//控制子进程
void ctrlSlaver(const vector<channel>& channels)
{int which = 0;//int cnt = 5;while(true){int select = 0;Menu();                              //手动控制cout << "please Enter@" << endl;cin >> select;if (select <= 0 || select >= 5) break;//select >0 &&select<5//1.选择任务//int cmdcode = rand() % tasks.size();//创造一个随机数，当作任务；int cmdcode = select-1;//2.选择进程(负载均衡：不可以一直给某个进程派发任务，其他进程都闲着，要均衡发展这个叫做负载均衡)1.使用随机数  2.轮转l来实现负载均衡//int processpos = rand() % channels.size();						//随机数//cout << "father say:" << "cmdcode:" << cmdcode << "already send to" << channels[processpos]._slaverid << endl;cout << "father say:" << "cmdcode:" << cmdcode << "already send to" << channels[which]._slaverid << endl;//轮转//3.发送任务//write(channels[processpos]._cmdfd, &cmdcode, sizeof(cmdcode));      //随机数版本write(channels[which]._cmdfd, &cmdcode, sizeof(cmdcode));			  //轮转which++;which %= channels.size();//cnt--;//sleep(1);}
}
void QuitProcess(const vector<channel> &channels)						//退出进程
{for (const auto& c : channels){close(c._cmdfd);}//sleep(5);for (const auto& c : channels) waitpid(c._slaverid, nullptr, 0);   //等待退出//sleep(5);
}int main()
{Loadtask(&tasks);							//加载任务参数；srand(time(nullptr) ^ getpid() ^ 1023);		//种一个随机数种子//再组织vector<channel> channels;//初始化-------bug?————画图就可以看出来，推荐画一下图！！！答：文件描述符随着创建子进程递增；InitProcessPool(&channels);					//初始化Debug(channels);							//测试//2.开始控制子进程ctrlSlaver(channels);						//控制子进程//3.清理收尾QuitProcess(channels);return 0;
}