c++ - 创建 3 个子进程并在指定秒数后退出它们

Question

输出应该是什么样子的图像:我的问题是我需要编写一个程序来接受 3 个进程的名称作为命令行参数。每个进程将运行如下秒数:(PID%10)*3+5 并终止。这 3 个子进程终止后，父进程将重新安排每个 child 。当所有子进程都被重新安排 3 次后，父进程将终止。我已经使用 fork 创建了三个 child ，但很难让他们按照特定标准退出？

using namespace std;

int main(){
    int i;
    int pid;
    for(i=0;i<3;i++) // loop will run n times (n=3)
    {
        if(fork() == 0)
        {
            pid = getpid();
            cout << "Process p" << i+1 << " pid:" << pid << " Started..." << endl;
            exit(0);
        }
    }
    for(int i=0;i<5;i++) // loop will run n times (n=3)
    wait(NULL);

  }

Answer 1

您可以使用sigtimedwait等待 SIGCHLD 或超时。

工作示例:

#include <cstdio>
#include <cstdlib>
#include <signal.h>
#include <unistd.h>

template<class... Args>
void start_child(unsigned max_runtime_sec, Args... args) {
    // Block SIGCHLD.
    sigset_t set;
    sigemptyset(&set);
    sigaddset(&set, SIGCHLD);
    sigprocmask(SIG_BLOCK, &set, nullptr);
    // Enable SIGCHLD.
    signal(SIGCHLD, [](int){});

    pid_t child_pid = fork();
    switch(child_pid) {
    case -1:
        std::abort();
    case 0: {
        // Child process.
        execl(args..., nullptr);
        abort(); // never get here.
    }
    default: {
        // paren process.
        timespec timeout = {};
        timeout.tv_sec = max_runtime_sec;
        siginfo_t info = {};
        int rc = sigtimedwait(&set, nullptr, &timeout);
        if(SIGCHLD == rc) {
            std::printf("child %u terminated in time with return code %d.\n", static_cast<unsigned>(child_pid), info.si_status);
        }
        else {
            kill(child_pid, SIGTERM);
            sigwaitinfo(&set, &info);
            std::printf("child %u terminated on timeout with return code %d.\n", static_cast<unsigned>(child_pid), info.si_status);
        }
    }
    }
}

int main() {
    start_child(2, "/bin/sleep", "/bin/sleep", "10");
    start_child(2, "/bin/sleep", "/bin/sleep", "1");
}

输出:

child 31548 terminated on timeout with return code 15.
child 31549 terminated in time with return code 0.