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Extension · App Inventor 2 中文网

...监听服务端返回的数据,再通过事件把结果切回主线程抛给积木层。 原作者与历史维护信息保留如下: Properties ConnectionState 返回当前连接状态。 返回 true 表示当前 Socket 已建立连接; 返回 ...
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C++实现一款简单完整的聊天室服务器+客户端 - C/C++ - 清泛网 - 专注C/C++及内核技术

...控制台读取一行数据 gets(buf); printf(">"); //发送给服务器 if(send(sock,buf,strlen(buf),0)==SOCKET_ERROR){ printf("发送数据失败!"); exit(-1); } } //清理套接字占用的资源 WSACleanup(); return 0; } DWORD WINAPI ReceiveThreadProc(...
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REDHAT 6.4 X64下ORACLE 11GR2静默安装 - 数据库(内核) - 清泛网 - 专注C/C++及内核技术

...认 SELINUX 和IPTABLES关闭 生产环境中防火墙功能就应该交给专门的硬件防火墙去解决。系统的功能越单一越好。 #vi /etc/selinux/config 确认里面的SELINUX字段的值是disabled,如果不是则改为disabled。 #service iptables stop 4、软件包...
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App Inventor 2 云数据库配置教程 - Firebase/云数据库1/自建后台 · App Inventor 2 中文网

...组件。本文将全面介绍App Inventor 2提供的云存储方案,并给出详细的配置步骤和实践代码。 一、云存储方案对比 App Inventor 2内置了多种云端数据存储组件,每种方案各有适用场景: 特性 云数...
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Win32汇编--使用MASM - C/C++ - 清泛网 - 专注C/C++及内核技术

...时候指定初始化值的,因为local伪指令只是简单地把空间给留出来,那么开始使用时它里面是什么值呢?和全局变量不一样,局部变量的初始值是随机的,是其他子程序执行后在堆栈里留下的垃圾,所以,对局部变量的值一定要...
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Defining an array of anonymous objects in CoffeeScript

... island205island205 1,6821616 silver badges2525 bronze badges 6 ...
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Java HashMap performance optimization / alternative

...et = very very bad. However if I turn the two arrays into a number in base 52 I am guaranteed to get a unique hash code for every object: public int hashCode() { // assume that both a and b are sorted return a[0] + powerOf52(a[1], 1) + powerOf52(b[0], 2) + powerOf52(b[1], 3) +...
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How many double numbers are there between 0.0 and 1.0?

... Java doubles are in IEEE-754 format, therefore they have a 52-bit fraction; between any two adjacent powers of two (inclusive of one and exclusive of the next one), there will therefore be 2 to the 52th power different doubles (i.e., 4503599627370496 of them). For example, that's th...
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vs2010编译boost若干问题解决 - 开源 & Github - 清泛网 - 专注C/C++及内核技术

...ttp: www.boost.org users download 上下载boost安装包,我下的是1.52.0版。按照说明,直接运行bootstr...首先说下环境,win7,vs2010。 先在http://www.boost.org/users/download/上下载boost安装包,我下的是1.52.0版。 按照说明,直接运行“bootstrap.bat...
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biggest integer that can be stored in a double

...in IEEE 64-bit doubles without losing precision. An IEEE 64-bit double has 52 bits of mantissa, so I think it's 253: 253 + 1 cannot be stored, because the 1 at the start and the 1 at the end have too many zeros in between. Anything less than 253 can be stored, with 52 bits explicitly stored in th...