final

What is a final variable in Java? For example: if I write final int temp; in function what is the meaning of the final keyword? Also, when would I want to use final variable (both as a class variable and as a function variable)? Why must variables in a synchronized block be declared final? Basically it just means you can't change the value. For instance variables, you have to assign any final variables once (and only once) in the constructor (or with a variable initializer). Synchronization is a pretty orthogonal concept. The primary reason for making a local variable final is so you can use

Is there way to map immutable Value objects like email address using JPA? @Immutable @Embeddable public final class EmailAddress { private final String value; public EmailAddress(String value) { this.value = value; } public String getValue() { return value; } @Override public boolean equals(Object o) { if (this == o) return true; if (o == null || getClass() != o.getClass()) return false; EmailAddress that = (EmailAddress) o; return value.equals(that.value); } @Override public int hashCode() { return value.hashCode(); } } Now I get exception on entity save org.hibernate.InstantiationException:

I could not find any references online about this. But just wanted to know if final fields in a class should always be static or is it just a convention. Based on my understanding of their uses, I feel that it is more of a logical thing to do than something that is imposed by the language. JB Nizet Of course not. They must be static if they belong to the class, and not be static if they belong to the instance of the class: public class ImmutablePerson { private static final int MAX_LAST_NAME_LENGTH = 255; // belongs to the type private final String firstName; // belongs to the instance private

This question already has an answer here: Why can not I add two bytes and get an int and I can add two final bytes get a byte? 2 answers Java - char, int conversions 4 answers The following segment of code issues a compile-time error. char c = 'c'; char d = c + 5; The error on the second line says, possible loss of precision required: char found: int The error message is based on the NetBeans IDE. When this character c is declared final like as follows. final char c = 'c'; char d = c + 5; The compiler-time error vanishes. It is unrelated to the case of final strings What does the final

final关键字 　　　　final就是一个关键字，表示最终的。 　　　　 　　　　final修饰的类 无法被继承；final 修饰的方法 不能重写（覆盖 ），但是可以 被继承 　　　　　　 　　　　　　 看一下示例，先是 final类 ： 在继承时直接报错 　　　　　　 　　　　　　 　　　　　　再是 final 修饰的 父类方法：类可以被继承，但是此时的方法重写时报错 　　　　　　 　　　　　　 　　　　　　 　　　　　　final修饰的变量一旦被赋值后 无法再次赋值 。 　　　　　　 　　　　　　这里就要说一下， final和static 一起用时是 定义一个常量，常量名所有字母都要大写 　　　　　　 final修饰的引用， 一旦指向某个对象 以后 不能再指向其他对象 　　　　　　　　先看看没有final修饰的： 　　　　　　　　 　　　　 　　　　　　　　再是有final修饰： 　　　　　　　　 　　　　　　 　　　　　　 　　 　　　　　　 来源： https://www.cnblogs.com/javaisbest/p/11600858.html

I'd like to keep my code compilable both on legacy C++ (C++ code using "NULL") and new C++11 standard (C++ code using "nullptr") I'm using GCC, but planning to recompile the whole codebase also for VS when I'll finish most important things. Should I expect both GCC and VS will do something like #define NULL nullptr Or Is better I'll do that myself (using of course a different name, where MY_LIB will be replaced by my library suffix)? #ifndef nullptr #define MY_LIB_NULL NULL #else #define MY_LIB_NULL nullptr #endif What I want to achieve is code that compiles regardless of wich C++11 features

1、final用于让虚函数不可被重写 struct B2 { virtual void f() final {} // final 函数 }; struct D2 : B2 { virtual void f() {} }; 如上代码是不可被编译过的 2、override 1.在函数比较多的情况下可以提示读者某个函数重写了基类虚函数（表示这个虚函数是从基类继承，不是派生类自己定义的）； 2.强制编译器检查某个函数是否重写基类虚函数，如果没有则报错。 来源： https://www.cnblogs.com/judes/p/11595365.html

加密 //cipher加密算法 function cipher ( str ) { try { const crypto = require ( 'crypto' ) ; /** * --创建 Cipher 实例。 不能使用 new 关键字直接地创建 Cipher 对象 * --crypto.createCipher,@param1 算法，@param2 密文，@param3 向量--可为""或省略 */ const cipher = crypto . createCipher ( 'aes128' , '冷月心' ) ; /** * update方法 * @param1 加密的数据 * @param2 数据编码格式，一般为utf8 * @param3 输出格式,一般为 'base64' 或者 'hex'，缺省返回Buffer */ let encrypted = cipher . update ( str , 'utf8' , 'hex' ) ; /** * final方法，返回加密后结果 * @param 返回值的字符编码 ,一般为 'base64' 或者 'hex'，缺省返回Buffer * -- 一旦 cipher.final() 方法被调用， Cipher 对象就不能再用于加密数据。 * -- 如果试图再次调用 cipher.final()，将会抛出一个错误。 */

This question already has an answer here: Does the Java 'final' keyword actually improve security? 12 answers I have read in Wikipedia here that: A final class cannot be subclassed. This is done for reasons of security and efficiency. I am wondering about the kind of security that final in Java can achieve? final does not provide any security whatsoever. It's silly to expect it to. It provides a kind of safety. Though your question is related to java, I'll point you to Marshall Cline's C++ FAQ which answers a similar question. http://www.parashift.com/c++-faq/encap-is-not-security.html [7.8]