Unix Network Programming Episode 88

文章介绍了Unix系统中inetddaemon如何解决早期服务启动问题,通过集中处理服务启动任务、减少进程数量和提供协议独立的服务,简化了服务器程序编写。它通过统一的接口接受来自TCP和UDP的客户端请求,提高了效率。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

‘inetd’ Daemon

On a typical Unix system, there could be many servers in existence, just waiting for a client request to arrive. Examples are FTP, Telnet, Rlogin, TFTP, and so on. With systems before 4.3BSD, each of these services had a process associated with it. This process was started at boot-time from the file /etc/rc, and each process did nearly identical startup tasks: create a socket, bind the server’s well-known port to the socket, wait for a connection (if TCP) or a datagram (if UDP), and then fork. The child process serviced the client and the parent waited for the next client request. There are two problems with this model:

1.All these daemons contained nearly identical startup code, first with respect to socket creation, and also with respect to becoming a daemon process (similar to our daemon_init function).
2.Each daemon took a slot in the process table, but each daemon was asleep most of the time.

The 4.3BSD release simplified this by providing an Internet superserver: the inetd daemon. This daemon can be used by servers that use either TCP or UDP. It does not handle other protocols, such as Unix domain sockets. This daemon fixes the two problems just mentioned:

1.It simplifies writing daemon processes since most of the startup details are handled by inetd. This obviates the need for each server to call our daemon_init function.
2.It allows a single process (inetd) to be waiting for incoming client requests for multiple services, instead of one process for each service. This reduces the total number of processes in the system.

Specifying the wait flag for a datagram service changes the steps done by the parent process. This flag says that inetd must wait for its child to terminate before selecting on this socket again. The following changes occur:

1.When fork returns in the parent, the parent saves the process ID of the child. This allows the parent to know when this specific child process terminates, by looking at the value returned by waitpid.
2.The parent disables the socket from future selects by using the FD_CLR macro to turn off the bit in its descriptor set. This means that the child process takes over the socket until it terminates.
3.When the child terminates, the parent is notified by a SIGCHLD signal, and the parent’s signal handler obtains the process ID of the terminating child. It reenables select for the corresponding socket by turning on the bit in its descriptor set for this socket.

‘daemon_inetd’ Function

We can call from a server we know is invoked by inetd.

#include "unp.h"
#include <syslog.h>

extern int daemon_proc;

void daemon_inetd(const char *pname, int facility)
{
    daemon_proc=1;
    openlog(pname, LOG_PID, facility);
}

daemon_inetd function: daemonizes process run by inetd

#include "unp.h"
#include <time.h>

int main(int argc, char **argv)
{
    socklen_t len;
    struct sockaddr *clientaddr;
    char buff[MAXLINE];
    time_t ticks;

    daemon_inetd(argv[0],0);

    clientaddr=Malloc(sizeof(struct sockaddr_storage));
    len=sizeof(struct sockaddr_storage);
    Getpeername(0,clientaddr, &len);
    err_msg("connection from %s", Sock_ntop(clientaddr, len));

    ticks=time(NULL);
    snprintf(buff, sizeof(buff), "%.24s\r\n", ctime(&ticks));
    Write(0,buff, strlen(buff));

    Close(0);
    return 0;
}

Protocol-independent daytime server that can be invoked by inetd

<think>好的,我现在需要处理用户关于强化学习中“episode reward”的定义和实现的问题。首先,我需要确保自己正确理解相关概念,并参考用户提供的资料来构建回答。 根据用户提供的引用,特别是引用[4],里面提到即时奖励和回报(return)的区别。即时奖励是执行动作后的立即反馈,而回报是回合中所有奖励的累积总和,是优化的目标。这里提到的“回报”应该对应英文中的“return”,而用户问的是“episode reward”,可能指的是整个回合的总奖励,也就是“return”。需要确认术语是否一致。 接下来,用户还提到了引用[3],其中讨论了信用分配问题,即确定哪些之前的动作导致了奖励。这可能与如何计算和分配episode reward中的各个奖励有关,尤其是在时间步骤较长的情况下。 定义部分,需要明确episode reward的定义,即在一个episode(从初始状态到终止状态的完整序列)中所有即时奖励的总和。可能还需要区分“episode reward”和“return”是否同义,或者是否有折扣因子应用。根据引用[4],回报是累积总和,可能考虑折扣,而用户的问题可能希望明确是否有折扣,或者是否只是简单相加。 实现方面,可能需要说明在代码中如何累加每个步骤的奖励。例如,在强化学习训练循环中,每个episode开始时将总奖励重置为0,然后在每个步骤累加即时奖励。如果有折扣,还需要应用折扣因子。 此外,用户提供的引用[1]提到了奖励、回报和值函数的区别,这可能帮助区分episode reward与其他概念。例如,值函数是期望回报,而episode reward是实际获得的回报总和。 需要考虑用户可能的混淆点,比如episode reward是否包含折扣,或者如何在实际算法中计算。同时,引用[3]中的信用分配问题可能影响如何设计算法来优化episode reward,例如使用策略梯度方法时需要考虑长期回报。 在生成相关问题的时候,需要围绕episode reward的相关概念,比如与step reward的区别、折扣因子的影响、实现时的代码结构,以及信用分配问题如何影响episode reward的使用。 现在,我需要整理这些思路,确保定义准确,引用正确,并给出清晰的实现示例。同时,检查是否有遗漏的关键点,比如是否区分了有限时域和无限时域的情况,或者是否在非折扣情况下讨论。 例如,在实现部分,可以展示一个简单的Python循环,在每个episode中累加奖励。同时,可能需要解释折扣因子gamma的作用,以及如何在代码中体现。但用户的问题可能更关注定义和基本实现,所以需要保持示例简洁。 最后,确保回答结构清晰,分点明确,引用标记正确,并生成相关问题帮助用户进一步学习。</think>### Episode Reward 的定义与实现 在强化学习中,**episode reward**(回合回报)指智能体在一个完整回合(episode)中获得的**所有即时奖励的累积总和**。它是评估智能体在单个回合中整体表现的核心指标,也是优化策略的直接目标[^4]。 #### 核心概念 1. **即时奖励(Immediate Reward)** 智能体在每一步执行动作后,环境反馈的瞬时奖励值,记为$r_t$。例如,在游戏中击中目标可能获得+10奖励。 2. **Episode Reward** 从回合开始到终止(如游戏胜利或失败),所有即时奖励的总和。计算公式为: $$ R = \sum_{t=0}^{T} r_t $$ 其中$T$为回合终止的时间步。若考虑折扣因子$\gamma$(平衡即时与未来奖励),则称为**折扣回报**: $$ R = \sum_{t=0}^{T} \gamma^t r_t $$ [^1][^4] 3. **与值函数的区别** 值函数(如$V(s)$)是**期望回报**,表示从状态$s$出发的长期预期收益;而episode reward是实际执行一个回合后获得的**具体数值**,用于训练时更新策略。 --- #### 实现方式 以游戏训练为例,实现episode reward的典型步骤如下: 1. **初始化累计变量** 每个回合开始时,将累计奖励`episode_reward`重置为0。 ```python episode_reward = 0 ``` 2. **循环执行动作并累加奖励** 在回合的每一步中,执行动作后获取即时奖励并累加: ```python state = env.reset() done = False while not done: action = agent.select_action(state) # 策略选择动作 next_state, reward, done, _ = env.step(action) episode_reward += reward # 累加即时奖励 state = next_state ``` 3. **策略优化** 使用回合结束后的`episode_reward`更新策略(如策略梯度方法): ```python # 假设使用蒙特卡洛策略梯度 optimizer.zero_grad() loss = -torch.log(probabilities) * episode_reward # 损失函数设计 loss.backward() optimizer.step() ``` --- #### 关键挑战 - **信用分配问题**:若回合后期获得高奖励,需确定哪些早期动作对此有贡献[^3]。 - **稀疏奖励**:若奖励仅在回合结束时给出(如围棋胜利),需通过值函数估计或稀疏奖励处理方法解决。 --- ### 相关问题 1. **如何通过折扣因子平衡即时与未来奖励?** 2. **信用分配问题如何影响深度强化学习的训练效率?** 3. **在稀疏奖励场景下,有哪些方法可以优化episode reward?** 4. **蒙特卡洛方法与时序差分方法在计算episode reward时有和区别?** : The difference between the reward, return, and value function : While the idea is quite intuitive, in practice there are numerous challenges... [^4]: 此外,需要区分即时奖励和回报...
评论 1
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值