release | 易学教程

Heroku docker spring boot image error 503 H14

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问题 I few days ago I was trying to deploy a docker image of a spring boot app into heroku. I have the following docker file: FROM openjdk:8-jdk-alpine LABEL maintainer="gabigarciagar@gmail.com" VOLUME /tmp COPY ./target/*.jar PetApp_Auth.jar CMD [ "java","-Xmx300m","-Xss512k","-XX:CICompilerCount=2","-Dfile.encoding=UTF-8","-Dspring.profiles.active=heroku","-Djava.security.egd=file:/dev/./urandom","-jar","/PetApp_Auth.jar" ] also in my application.properties for a heroku profile i have: spring

使用windbg调试release程序

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简介：最近经常会碰到客户那边出现各种奇怪的问题，由于日志不可能完全记录运行信息，又或者日志回滚导致记录信息丢失等，无法定位bug出现位置，这时你非常希望自己的能够单步调试。下面是vs2008+windbg研究的结果，可能存在不完善的地方，希望牛人不吝赐教，互相学习。<br/> 补充：一般写完程序之后，自己会写一些测试用例来测试自己写的程序，进行完单元测试之后会交由测试部进行测试，但是不能够保证所有的bug都会被测试出来，多数会由客户电脑负责的环境导致程序出现一些莫名其妙的问题，废话不多说了。<br/> 调试环境测试环境：windows 7<br/> 开发工具：vs2008, windbg6.3.9600<br/> ##调试代码## #include <stdio.h> void TestFunc(int nVar, char *pStr) { printf("juest print nVar:%d, pStr:%s\n", nVar, pStr ? pStr:"null"); } void main() { TestFunc(1, "hello world!"); } ##vs2008 release模式设置## /Zi 该命令是生成PDB文件中包含变量名和函数名，不会影响到优化选项(生成代码时vs release版会对代码进行优化) ![在此输入图片描述][1] /01

Qt 5.0.2 静态编译的base release库

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编译参数如下: configure -release -static -platform win32-msvc2010 -qt-zlib -qt-libpng -qt-libjpeg -qt-sql-sqlite -plugin-sql-sqlite -qt-style-windows -qt-style-windowsxp -qt-style-windowsvista -nomake demos -nomake examples -nomake docs -mp -no-icu -opengl desktop -angle 关闭了ICU. 编译过程中opengl环节出现一些问题.如果程序中不用到opengl应该是没问题的.. http://pan.baidu.com/share/link?shareid=450973&uk=3271863450 来源： oschina 链接： https://my.oschina.net/u/179574/blog/127447

Unable to run qt-creator executable outside ide compiled in release mode

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问题 After switch to release mode to build a small project I have, when I try run it from inside the qt-creator ide, all goes fine. But when I go to the folder build-<project_name>-Desktop_Qt_5_14_1_MinGW_64_bit- Release and try run the executable generated in this directory, I got this error: Anyone knows what the problem here? If it was some missing dll, I supose it will specify what dll was missing, right? Or I am mistaken? update After run windeployqt , this command found dependencies for my

SpringCloud和SpringBoot的详细版本说明

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spring boot 和cloud版本说明 springcloud版本号查看： https://spring.io/projects/spring-cloud#learn 大版本说明大版本： Spring Boot Spring Cloud 关系 1.2.x Angel版本(天使) 兼容Spring Boot 1.2.x 1.3.x Brixton版本(布里克斯顿) 兼容Spring Boot 1.3.x，也兼容Spring Boot 1.4.x 1.4.x Camden版本(卡姆登) 兼容Spring Boot 1.4.x，也兼容Spring Boot 1.5.x 1.5.x Dalston版本(多尔斯顿) 兼容Spring Boot 1.5.x，不兼容Spring Boot 2.0.x 1.5.x Edgware版本(埃奇韦尔) 兼容Spring Boot 1.5.x，不兼容Spring Boot 2.0.x 2.0.x Finchley版本(芬奇利) 兼容Spring Boot 2.0.x，不兼容Spring Boot 1.5.x 2.1.x Greenwich版本(格林威治) 实际开发版本关系 spring-boot-starter-parent spring-cloud-dependencies 版本号发布日期版本号发布日期 1.5.2.RELEASE

ubuntu的版本生命周期

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Releases https://wiki.ubuntu.com/Releases List of releases Ubuntu Website release cycle page Current Version Code name Docs Release End of Standard Support End of Life Ubuntu 19.10 Eoan Ermine Release Notes October 17, 2019 July, 2020 July, 2020 Ubuntu 18.04.4 LTS Bionic Beaver Changes February 12, 2020 April 2023 April 2028 Ubuntu 18.04.3 LTS Bionic Beaver Changes August 8, 2019 April 2023 April 2028 Ubuntu 18.04.2 LTS Bionic Beaver Changes February 15, 2019 April 2023 April 2028 Ubuntu 18.04.1 LTS Bionic Beaver Changes July 26, 2018 April 2023 April 2028 Ubuntu 18.04 LTS Bionic Beaver

死锁

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死锁状态 from multiprocessing import Process from multiprocessing import Lock,RLock def func(lock): lock.acquire() # 持有锁 print('子进程成功拿锁') lock.acquire() # 持有锁 print('子进程再次成功拿锁') lock.release() # 使用release() 释放资源 lock.release() return def main(): l = RLock() # 使用Lock()会发生死锁，使用RLock()避免死锁 p = Process(target=func,args=(l,)) p.start() p.join() if __name__ == '__main__': main() 比较两个进程认识死锁 from multiprocessing import Process from multiprocessing import Lock,RLock import time def a(l1,l2): l1.acquire() # 持有l1锁 print('a进程拿到了l1锁') time.sleep(3) # 加时间体现死锁状态 l2.acquire() # 持有l2锁 print('a进程拿到了l2锁') print(

Distinguish between key press and release of MIDI piano input

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问题 I am about to make a little program for a school project that is supposed to recognize the chords that are being played via a MIDI piano input (that's just one part of it). At the moment I got so far that for each press and each release of a key on the midi keyboard I get an object of the class ShortMessage . My question: How do I figure out if the key has been pressed or released? In each case, press and release, the static variable NOTE_OFF does contain the value 128, the variable NOTE_ON

Visual Studio - Debug vs Release

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问题 I built a windows service, targeted for .NET 2.0 in VS 2008. I run it as a console app to debug it. Console app is working great. I put it on my local computer as a service, compiled in debug mode, still working great. I'm ready to release now, and suddenly, when I set it to release mode, the service compiles and installs, but nothing happens. (No code in service is running at all). I realize that the release vs debug mode are property configuration settings, but it seems that in release mode

Visual Studio - Debug vs Release

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