static

Why can't constructor be declared as static in c++?

由 前提是你 提交于 2020-01-13 08:53:15
问题 I have recently finished reading the 1st Vol. of Thinking in C++ by Bruce Eckel and have now turned to applying the knowledge to some practical use. I was recently working with static member functions and tried making the constructor static, for which the compiler was unhappy. I checked for the reason in the book but couldn't find any. Can anyone explain why? P.S.: After seeing some responses, I would like to mention that the confusion arises from my knowledge that C# (and Java) allows

Why can't constructor be declared as static in c++?

由 南笙酒味 提交于 2020-01-13 08:53:06
问题 I have recently finished reading the 1st Vol. of Thinking in C++ by Bruce Eckel and have now turned to applying the knowledge to some practical use. I was recently working with static member functions and tried making the constructor static, for which the compiler was unhappy. I checked for the reason in the book but couldn't find any. Can anyone explain why? P.S.: After seeing some responses, I would like to mention that the confusion arises from my knowledge that C# (and Java) allows

【单片机笔记】上海移远公司NB-IOT模组 BC26 使用STM32 AT命令实现连接阿里云数据上传和下载

由 回眸只為那壹抹淺笑 提交于 2020-01-13 08:51:55
前言 在调试之前看这个数据手册一脸懵，特别是MQTT部分还是独立的，这个和前接触到的上海合宙的模块多少有点出处。另外就是那个AT命令的传入参数也是一脸懵，后来发现BC26的模块好像把MQTT部分单独的做成了支持阿里云服务器的功能。接触过阿里云的设备对接相比都知道，阿里云要求的是一机一密或者一型一密，这个在对于简单的成本低廉的MCU来说无疑是一个很大的考验。而BC26这块还是做的非常友好的，在MQTT部分只需要传入产品对应设备下的三元组即可，无需经过哈希算法计算密钥。 先上图： 连接及上传部分： 数据下载部分： 基本上通信也是非常稳定的，不过我这个地方信号贼差，以模块满格31来算，这里测试才有9，10的样子。 整体的应用层思路是这样的： 1、单片机控制模块自检 2、单片机控制模块连接阿里云 3、单片机控制模块定时发送数据并接收下发数据 看上面三个步骤虽然简洁明了，单是要做好单片机的应用底层也不是那么简单，为此我专门谢了一个对应BC26传输机制的地层代码，可以检测BC26模块的状态，连接状态，断线重连、超时复位等机制。详细请看下文代码： BC26底层驱动部分： C文件： #include "fy_bc26.h" static void ResetModule(void); static void CheckModule(void); static void SetCFUN(void);

Java - Get reference to a static class using reflection

由 大城市里の小女人 提交于 2020-01-13 08:36:10
问题 In Java, is it possible to access an instance of a static class (nested) using reflection? Supposing I have the following 2 classes defined in the package Package1.SubPackage.SubSubPackage: public class MyMainClass { public static class SalesObjectGrouper1 { public static final GrouperContext CONTEXT = new GrouperContext("MyDate"); } private static class SalesObjectGrouper2 { public static final GrouperContext CONTEXT = new GrouperContext("MyDate"); } } If I run the following code: try { xyz

Private static variables in php class

由 本秂侑毒 提交于 2020-01-13 08:33:32
问题 I have a few classes that are often run through var_dump or print_r . Inside these classes I have a few variables that are references to other, rather large objects that only ever has one instance of each and are only used inside the classes (outside the classes have their own reference to these classes) I do not wish these classes printed in the output, so I have declared them as private static which is working fine. But my IDE (PHPstorm) is flicking up an error-level alert with Member has

六、关键字

由 别等时光非礼了梦想. 提交于 2020-01-13 05:43:26
4.关键字： （1）this：this是一个关键字，存储了当前对象的内存地址 -->this对象，用来访问本类自己的属性和方法。 -->构造器中，给属性set值。 【1】 使用本类的属性 ，this.属性名, this可以省略不写 【2】 调用本类的方法 this.方法名() ,this可以省略不写 【3】 调用本类的构造方法, 要求必须是构造方法中的第一句代码 this()—》调用本类的无参构造方法 this(实参列表)调用本类的带参构造方法 【4】 当局部变量与成员变量名称相同时，this代表成员变量 （2）continue：跳出本次循环继续下一次循环； break：跳出循环体，继续执行循环外的函数体；在switch...case中终止与下一个case的比较； （3） return: 跳出整个函数体，函数体后面的部分不再执行。 （4）final关键字： 修饰类：final修饰的类不能被继承，其中的属性可以自由定义类型，但是其中的方法默认被final，不能修改。（除非真的不想其被继承，出于安全问题，一般不设置为final）； 修饰方法：如果只有在想明确禁止 该方法在子类中被覆盖的情况下才将方法设置为final的。 修饰属性：基本数据类型：值不能被修改，固定值。引用数据类型：在栈内存中的存储的地址不变， 对于一个final变量，如果是基本数据类型的变量

Java Static confusion

由 感情迁移 提交于 2020-01-13 05:07:08
问题 I'm working with Java; I have worked with C++ before. I am thinking about static usage in Java. If I create static methods and variables in the class, Why can I access them through the object also? Example: class Test{ static int count=0; int id; static void updatec(){ count++ } } class TestMain { public static void main(String args[]) { Test.count=1; Test t = new Test(); t.count=5; // Valid WHY ????? } } Why this is allowed? Java's site says we should not use obj.static method/variable. Why

The meaning of static in C++

由 假装没事ソ 提交于 2020-01-13 03:53:05
问题 I thought I was fairly good with C++, it turns out that I'm not. A previous question I asked: C++ const lvalue references had the following code in one of the answers: #include <iostream> using namespace std; int& GenX(bool reset) { static int* x = new int; *x = 100; if (reset) { delete x; x = new int; *x = 200; } return *x; } class YStore { public: YStore(int& x); int& getX() { return my_x; } private: int& my_x; }; YStore::YStore(int& x) : my_x(x) { } int main() { YStore Y(GenX(false)); cout

C++: Do static primitives become invalid at program exit?

由 為{幸葍}努か 提交于 2020-01-13 03:18:06
问题 Assume I have a function like this: MyClass &MyFunction(void) { static MyClass *ptr = 0; if (ptr == 0) ptr = new MyClass; return MyClass; } The question is at program exit time, will the ptr variable ever become invalid (i.e. the contents of that ptr are cleaned up by the exiting process)? I realize that this function leaks, but it is only an example for simplicity. The same question also applies to other primitives besides pointers as well. How about if I have a static integer, does the

4.C++ 学习 进一步补充(static,单例设计,类型模板)

由 丶灬走出姿态 提交于 2020-01-13 01:29:43
static 在类中 static数据 例子: 在银行系统中,有一万个人,每个人都有不同的账户,但是利率是唯一的所有的人每年的利率相同,我们就把利率这个设置为静态数据, static中的数据只初始化一次 static 函数 无法访问类所创建出的数据,无法对创建出的数据进行操作 只能对static数据进行操作 调用static函数的方法 通过类名加函数名的方式使用 Class::Func() 单例设计模式 class A { public : static A & getInstance ( ) { return a ; } //只能通过getInstance去取,取到的函数是已经创建好的 private : A ( ) ; A ( const A & rhs ) ; //构造函数在私有函数中 static A a ; } 类型模板 template < typename T > //模板,不用把类型写死,到时候再指定 class complex { public : complex ( T r = 0 , T i = 0 ) : re ( r ) , im ( i ) { } complex & operator + = ( const complex & ) ; T real ( ) const { return re ; } T imag ( ) const { return
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