c++ - 用c++进行fork()并创建父对象的4个子代

我在fork（）ing并创建同时运行的4个子进程（忙碌的工作），但是父级需要等待所有子级退出。
我觉得我可以用更好的方法编写代码。.有什么建议吗？看起来还好吗？ （我必须保持2个函数和main的这种格式）

using namespace std;

void simulateBusyWork(char, pid_t);
pid_t performFork();

int main()
{
    srand(time(NULL));
    performFork();
    return 0;
}

pid_t performFork() {

    pid_t pid;

    //create four child processes
    for (int i = 0; i < 4; ++i) {
        pid = fork();
        if (pid<0) {
            continue;
        } else if (pid == 0) {
            break;
        }
    }

    //fork failed
    if (pid < 0) {
        fprintf(stderr, "Fork failed./n");
        return 1;
    }

    //child process do busy work
    else if (pid == 0) {
        cout << "A child process with PID " << getpid() << " was created." << endl;
        pid = getpid();
        simulateBusyWork('C', pid);
        _exit(0);
    }

    //parent process do busy work
    else if(pid > 0){
        simulateBusyWork('P', getpid());

        //parent process waits for children to exit
        int status, exited;
        //print which child has exited
        for (int i = 0; i < 4; i++) {
            exited = waitpid(-1, &status, 0);
            cout << "P" << getpid() << ": Process " << exited << " has exited." << endl;
        }
    }



    return pid;
}

void simulateBusyWork(char ch, pid_t pid) {

    for (int i=0 ; i < 4; i++) {
        cout << ch << pid << ": " << i << endl;
        //if (i < 4)
            sleep(1);
    }
    cout << "Process " << ch << pid << " has finished its work." << endl;
}

最佳答案

man waitpid：


  呼叫wait(&status)是等效的
         至：

       waitpid(-1, &status, 0);

因此您可以简化一下。