Delphi 使用ZMQ2.2.0.0版本，pub端发布数据，sub连接后内存一直增长 5C

pub和sub在不同电脑上，sub不接收数据，只connect还是一直在增长，最后释放 zmq上下文，socket 内存也会降下来!

1个回答

LiYuStudio 最后 重启电脑 重新测试 内存就不增长了

The troubles of lmy 程序的实现
Problem Description Lmy is the lovely cousin of zmq, and she’s a junior high school student. One day she came back from school. Zmq: What did you learn in your class today? Lmy: Similar triangle. Zmq: Well, fine, have you mastered it? Lmy: Sure, that’s a piece of cake. Zmq: OK, well, let me test you to see whether you have thoroughly understood the conception of being similar. I’ll draw two Arbitrary Polygons, and you have to tell me whether they are similar or not. If you are right, I’ll take you out. Lmy: …… It seems that little lmy got into troubles. Can you help her out by designing a program? Input There are several test cases in the input. The first line of each case is an n (n <= 300), indicating that each polygon has n points. And in the following 2n lines, each line represents a point. The first n point represents the first polygon in anti-clockwise order, and so the second. Please note that all the polygons are simple. You can try to rotate them, then tell whether they are similar or not. Output For each case, output “Yes” or “No”, showing whether they are similar. Sample Input 3 0 0 1 1 0 1 0 0 2 2 0 2 3 0 0 1 1 0 1 0 0 2 2 0 1 Sample Output Yes No

mq: connect exception: WRAP_EXCEPTION
An error ocurred while starting the kernel

zmq发送几次数据后出现严重错误

The troubles of lmy C程序的编写技术
Problem Description Lmy is the lovely cousin of zmq, and she’s a junior high school student. One day she came back from school. Zmq: What did you learn in your class today? Lmy: Similar triangle. Zmq: Well, fine, have you mastered it? Lmy: Sure, that’s a piece of cake. Zmq: OK, well, let me test you to see whether you have thoroughly understood the conception of being similar. I’ll draw two Arbitrary Polygons, and you have to tell me whether they are similar or not. If you are right, I’ll take you out. Lmy: …… It seems that little lmy got into troubles. Can you help her out by designing a program? Input There are several test cases in the input. The first line of each case is an n (n <= 300), indicating that each polygon has n points. And in the following 2n lines, each line represents a point. The first n point represents the first polygon in anti-clockwise order, and so the second. Please note that all the polygons are simple. You can try to rotate them, then tell whether they are similar or not. Output For each case, output “Yes” or “No”, showing whether they are similar. Sample Input 3 0 0 1 1 0 1 0 0 2 2 0 2 3 0 0 1 1 0 1 0 0 2 2 0 1 Sample Output Yes NoThe troubles of lmy
zeromq 服务端多线程在数据量大一点的情况下出现问题,跪求大神解答
zeromq 服务端开启了20个线程,客服端会发送6000个请求过来; 在处理1000个请求的时候,速度很快,一口气就跑完了, 在处理一千到三千的请求的时候,中间总是停一会,跑一会, 到了三千以后停的时间越来越长,到四千左右就不动了. 有时候会报"gc开销过大",调了gc之后.还是没有多大的作用, 但是呢:当我把服务端的线程只开启一个的时候,就能顺利跑完, 下面是服务端的代码 ``` /** * 多线程 NLP 服务器 */ public class mtserver { /*默认为20个NLP线程*/ static Integer WORKERNUM = 20; private static class HeronlpWorker extends Thread { private ZContext context; private MyUncaughtExceptionhandler myUncaughtException; public void setMyUncaughtException(MyUncaughtExceptionhandler myUncaughtException) { this.myUncaughtException = myUncaughtException; } private HeronlpWorker(ZContext context) { this.context = context; } @Override public void run() { Socket socket = context.createSocket(SocketType.REP); myUncaughtException.setSkt(socket); socket.connect("inproc://workers"); HeroNLP heroNLP = new HeroNLP(); String result = null; String msgType = ""; while (true) { try { //接收命令 String command = socket.recvStr(0).trim(); System.out.println(Thread.currentThread().getName() + " Received request: [" + command + "]"); String content = null; switch (command){ //分词 case "SEGMENT": content = socket.recvStr(0).trim(); List<Term> terms = NLP.segment(content); result = listToString(terms, " "); break; default: while (socket.hasReceiveMore()){ socket.recvStr(0).trim(); } result = "FAILED: 未匹配到命令!"; break; } } catch (Exception e) { result = "FAILED:" + e.getMessage(); } // 给客户端返回结果(C字符串) System.out.println("是否有异常: " + result.contains("FAILED:")); if(result.contains("FAILED:")){ msgType = "FAILED"; } socket.send(msgType, 2); socket.send(result, 0); } } } public static void main(String[] args) { try (ZContext context = new ZContext()) { //前端--router Socket clients = context.createSocket(SocketType.ROUTER); boolean bind = clients.bind("tcp://*:9002"); //后端--dealer Socket workers = context.createSocket(SocketType.DEALER); workers.bind("inproc://workers"); //后端启动20个服务线程 for (int thread_nbr = 0; thread_nbr < WORKERNUM; thread_nbr++) { Thread nlpworker = new HeronlpWorker(context); MyUncaughtExceptionhandler myUncaughtExceptionhandler = new MyUncaughtExceptionhandler(); ((HeronlpWorker) nlpworker).setMyUncaughtException(myUncaughtExceptionhandler); nlpworker.setUncaughtExceptionHandler(myUncaughtExceptionhandler); nlpworker.start(); } ZMQ.proxy(clients, workers, null); } } ```
ZMQ通讯，在多对多模式中，如何指定地址发送？

go get zeromq4.0.5出现libzmq.pc错误

Problem Description Lmy is the lovely cousin of zmq, and she’s a junior high school student. One day she came back from school. Zmq: What did you learn in your class today? Lmy: Similar triangle. Zmq: Well, fine, have you mastered it? Lmy: Sure, that’s a piece of cake. Zmq: OK, well, let me test you to see whether you have thoroughly understood the conception of being similar. I’ll draw two Arbitrary Polygons, and you have to tell me whether they are similar or not. If you are right, I’ll take you out. Lmy: …… It seems that little lmy got into troubles. Can you help her out by designing a program? Input There are several test cases in the input. The first line of each case is an n (n <= 300), indicating that each polygon has n points. And in the following 2n lines, each line represents a point. The first n point represents the first polygon in anti-clockwise order, and so the second. Please note that all the polygons are simple. You can try to rotate them, then tell whether they are similar or not. Output For each case, output “Yes” or “No”, showing whether they are similar. Sample Input 3 0 0 1 1 0 1 0 0 2 2 0 2 3 0 0 1 1 0 1 0 0 2 2 0 1 Sample Output Yes No
tensorflow的InvalidArgumentError报错问题（placeholder）

The troubles of lmy
Problem Description Lmy is the lovely cousin of zmq, and she’s a junior high school student. One day she came back from school. Zmq: What did you learn in your class today? Lmy: Similar triangle. Zmq: Well, fine, have you mastered it? Lmy: Sure, that’s a piece of cake. Zmq: OK, well, let me test you to see whether you have thoroughly understood the conception of being similar. I’ll draw two Arbitrary Polygons, and you have to tell me whether they are similar or not. If you are right, I’ll take you out. Lmy: …… It seems that little lmy got into troubles. Can you help her out by designing a program? Input There are several test cases in the input. The first line of each case is an n (n <= 300), indicating that each polygon has n points. And in the following 2n lines, each line represents a point. The first n point represents the first polygon in anti-clockwise order, and so the second. Please note that all the polygons are simple. You can try to rotate them, then tell whether they are similar or not. Output For each case, output “Yes” or “No”, showing whether they are similar. Sample Input 3 0 0 1 1 0 1 0 0 2 2 0 2 3 0 0 1 1 0 1 0 0 2 2 0 1 Sample Output Yes No
zeromq socket网络传递结构体

Java学习的正确打开方式

linux系列之常用运维命令整理笔录

Python十大装B语法
Python 是一种代表简单思想的语言，其语法相对简单，很容易上手。不过，如果就此小视 Python 语法的精妙和深邃，那就大错特错了。本文精心筛选了最能展现 Python 语法之精妙的十个知识点，并附上详细的实例代码。如能在实战中融会贯通、灵活使用，必将使代码更为精炼、高效，同时也会极大提升代码B格，使之看上去更老练，读起来更优雅。 1. for - else 什么？不是 if 和 else 才

2019年11月中国大陆编程语言排行榜
2019年11月2日，我统计了某招聘网站，获得有效程序员招聘数据9万条。针对招聘信息，提取编程语言关键字，并统计如下： 编程语言比例 rank pl_ percentage 1 java 33.62% 2 c/c++ 16.42% 3 c_sharp 12.82% 4 javascript 12.31% 5 python 7.93% 6 go 7.25% 7

JDK12 Collectors.teeing 你真的需要了解一下

SQL-小白最佳入门sql查询一

【图解经典算法题】如何用一行代码解决约瑟夫环问题

“狗屁不通文章生成器”登顶GitHub热榜，分分钟写出万字形式主义大作

GitHub标星近1万：只需5秒音源，这个网络就能实时“克隆”你的声音

《程序人生》系列-这个程序员只用了20行代码就拿了冠军

11月8日，由中国信息通信研究院、中国通信标准化协会、中国互联网协会、可信区块链推进计划联合主办，科技行者协办的2019可信区块链峰会将在北京悠唐皇冠假日酒店开幕。 　　区块链技术被认为是继蒸汽机、电力、互联网之后，下一代颠覆性的核心技术。如果说蒸汽机释放了人类的生产力，电力解决了人类基本的生活需求，互联网彻底改变了信息传递的方式，区块链作为构造信任的技术有重要的价值。 　　1...

【技巧总结】位运算装逼指南

【管理系统课程设计】美少女手把手教你后台管理
【文章后台管理系统】URL设计与建模分析+项目源码+运行界面 栏目管理、文章列表、用户管理、角色管理、权限管理模块（文章最后附有源码） 1. 这是一个什么系统? 1.1 学习后台管理系统的原因 随着时代的变迁，现如今各大云服务平台横空出世，市面上有许多如学生信息系统、图书阅读系统、停车场管理系统等的管理系统，而本人家里就有人在用烟草销售系统，直接在网上完成挑选、购买与提交收货点，方便又快捷。 试想，若没有烟草销售系统，本人家人想要购买烟草，还要独自前往药...
4G EPS 第四代移动通信系统