From e6b0d1661d7b608dbcb247ce5735d505cefe1247 Mon Sep 17 00:00:00 2001
From: Sduby <supersduby@gmail.com>
Date: Mon, 14 Feb 2022 23:09:28 +0800
Subject: [PATCH] hw05

---
 main.cpp | 42 +++++++++++++++++++++++++++++++++++-------
 1 file changed, 35 insertions(+), 7 deletions(-)

diff --git a/main.cpp b/main.cpp
index f4ecab8..60f1bd8 100644
--- a/main.cpp
+++ b/main.cpp
@@ -1,6 +1,9 @@
 // 小彭老师作业05：假装是多线程 HTTP 服务器 - 富连网大厂面试官觉得很赞
+#include <chrono>
 #include <functional>
 #include <iostream>
+#include <mutex>
+#include <shared_mutex>
 #include <sstream>
 #include <cstdlib>
 #include <string>
@@ -15,12 +18,14 @@ struct User {
 };
 
 std::map<std::string, User> users;
-std::map<std::string, long> has_login;  // 换成 std::chrono::seconds 之类的
+std::map<std::string, std::chrono::seconds> has_login;  // 换成 std::chrono::seconds 之类的
+std::shared_mutex mtx_users, mtx_login;
 
 // 作业要求1：把这些函数变成多线程安全的
 // 提示：能正确利用 shared_mutex 加分，用 lock_guard 系列加分
 std::string do_register(std::string username, std::string password, std::string school, std::string phone) {
     User user = {password, school, phone};
+    std::unique_lock grd(mtx_users);
     if (users.emplace(username, user).second)
         return "注册成功";
     else
@@ -29,13 +34,18 @@ std::string do_register(std::string username, std::string password, std::string
 
 std::string do_login(std::string username, std::string password) {
     // 作业要求2：把这个登录计时器改成基于 chrono 的
-    long now = time(NULL);   // C 语言当前时间
-    if (has_login.find(username) != has_login.end()) {
-        int sec = now - has_login.at(username);  // C 语言算时间差
-        return std::to_string(sec) + "秒内登录过";
+    auto now = std::chrono::duration_cast<std::chrono::seconds>
+                    (std::chrono::steady_clock::now().time_since_epoch());
+    {
+        std::unique_lock grd(mtx_login);
+        if (has_login.find(username) != has_login.end()) {
+            auto sec = now - has_login.at(username);  // C 语言算时间差
+            return std::to_string(sec.count()) + "秒内登录过";
+        }
+        has_login[username] = now;
     }
-    has_login[username] = now;
 
+    std::shared_lock grd(mtx_users);
     if (users.find(username) == users.end())
         return "用户名错误";
     if (users.at(username).password != password)
@@ -44,7 +54,13 @@ std::string do_login(std::string username, std::string password) {
 }
 
 std::string do_queryuser(std::string username) {
-    auto &user = users.at(username);
+    std::shared_lock grd(mtx_users);
+    auto found = users.find(username);
+    if (found == users.end()) 
+        return "用户名不存在\n";
+    auto &user = found->second;
+    grd.unlock();
+
     std::stringstream ss;
     ss << "用户名: " << username << std::endl;
     ss << "学校:" << user.school << std::endl;
@@ -54,10 +70,21 @@ std::string do_queryuser(std::string username) {
 
 
 struct ThreadPool {
+    std::vector<std::thread> pool;
+
     void create(std::function<void()> start) {
         // 作业要求3：如何让这个线程保持在后台执行不要退出？
         // 提示：改成 async 和 future 且用法正确也可以加分
+
+        // A: 维护一个全局线程池，全局的tpool销毁时调用析构函数循环join()
         std::thread thr(start);
+        pool.push_back(std::move(thr));
+    }
+
+    ~ThreadPool() {
+        for(auto &thr: pool) {
+            thr.join();
+        }
     }
 };
 
@@ -85,5 +112,6 @@ int main() {
     }
 
     // 作业要求4：等待 tpool 中所有线程都结束后再退出
+    // 全局的tpool析构时自动join()
     return 0;
 }