instanceof

Java abstract method with abstract parameter and inheritance

£可爱£侵袭症+ 提交于 2019-12-05 20:44:23
I recently fumbled into a problem with an API and an implementation where the following type of code appeared: The API is an abstract class: public abstract class A { public A sum(A a) { System.out.println("A.sum(A) called"); return null; } } The implementation is a simple class: public class B extends A { public B sum(B b) { System.out.println("B.sum(B) called"); return null; } } When it comes to using it I write: public class Main { public static void main(String args[]) { B b = new B(); A basa = new B(); b.sum(b); basa.sum(b); basa.sum(basa); } } Which results in: % java Main B.sum(B)

JSF EL: instanceof reserved but not yet implemented?

大兔子大兔子 提交于 2019-12-05 18:10:32
问题 I've found the instanceof operator in JSF EL, but it throws an exception when used. It's obviously reserved but not implemented? When will it (probably) be available, if not already in a newer version than JSF 1.2, which I'm currently using? 回答1: The keyword instanceof is indeed reserved in the EL (see here), but it is still not used in the latest version of EL (2.2), used in JSF 2.x. 来源: https://stackoverflow.com/questions/5549999/jsf-el-instanceof-reserved-but-not-yet-implemented

Generics and instanceof - java

僤鯓⒐⒋嵵緔 提交于 2019-12-05 15:06:34
OK this is my class, it encapsulates an object, and delegates equals and to String to this object, why I can´t use instance of??? public class Leaf<L> { private L object; /** * @return the object */ public L getObject() { return object; } /** * @param object the object to set */ public void setObject(L object) { this.object = object; } public boolean equals(Object other) { if(other instanceof Leaf<L>) //--->ERROR ON THIS LINE { Leaf<L> o = (Leaf<L>) other; return this.getObject().equals(o.getObject()); } return false; } public String toString() { return object.toString(); } } how can I get

In jshell-11, why does a redeclared reference variable that resets to null still have a type?

北战南征 提交于 2019-12-05 14:15:27
问题 When redeclaring Integer 'a' in line 33, why does jshell show the reference variable as an instance of Integer (refer to lines 38 & 39)? After the redeclaration, line 34 shows that 'a' is set to null. When 'a' is declared in line 6 but not given a value, or reset to null in line 22, 'a' is not considered an instance of Integer. I would expect that when the reference variable is redeclared, since its value is null, that it would not be an instance of a type; however, that is not the case. 01:

“illegal generic type of instanceof” when using instanceof on an inner class type?

a 夏天 提交于 2019-12-05 13:50:01
I coded in NetBeans something like this: public class Grafo<V, E> { class Par { int a, b; Par(int a, int b) { this.a = a; this.b = b; } @Override public boolean equals(Object ob) { if(ob instanceof Par) { Par p = (Par)ob; return this.a==p.a && this.b==p.b; } return false; } } //stuff... } //end of class Grafo The error is in the method equals() from inner class "Par". NetBeans says that the error is "illegal generic type of instanceof". The error is in the line below. if(ob instanceof Par) { What is the cause of the error ? Try ob instanceof Grafo<?,?>.Par I think that the compiler thinks that

详解PHP中instanceof关键字及instanceof关键字有什么作用

为君一笑 提交于 2019-12-05 12:38:57
来源: https://www.jb51.net/article/74409.htm PHP5的另一个新成员是instdnceof关键字。使用这个关键字可以确定一个对象是类的实例、类的子类,还是实现了某个特定接口,并进行相应的操作。在某些情况下,我们希望确定某个类是否特定的类型,或者是否实现了特定的接口。instanceof操作符非常适合完成这个任务。instanceof操作符检查三件事情:实例是否某个特定的类型,实例是否从某个特定的类型继承,实例或者他的任何祖先类是否实现了特定的接口。例如,假设希望了解名为manager的对象是否为类Employee的实例: $manager = new Employee(); … if ($manager instanceof Employee) echo "Yes"; 有两点值得注意。首先,类名没有任何定界符(引号)。使用定界符将导致语法错误。其次,如果比较失败,脚本将退出执行。instanceof关键字在同时处理多个对象时特别有用。例如,你可能要重复地调用某个函数,但希望根据对象类型调整函数的行为。可以使用case语句和instanceof关键字来实现这个目标。 来源: https://www.cnblogs.com/laijinquan/p/11925440.html

JS常用关键字总结

China☆狼群 提交于 2019-12-05 11:03:44
in: 案例1、遍历对象: for(key in obj) { console.info( key+":"+obj[key]; ) }; 案例2、判断对象中是否有属性: "name" in obj 案例3、判断数组是否有此下标: 3 in ['a','b','c','d'] typeof: typeof可以判断一个值类型,对变量或值调用typeof运算符将返回下列字符串: Undefined:"undefined" 布尔:"boolean" 整数、浮点:"number" 字符串:"string" 函数:"function":如果变量是个函数 ( 判断事件是否被注册: if(typeof window.onload=="function"){alert("yes")} ) 除了基本类型和function,其他类型都返回:"object" instanceof: typeof可以用来检测基础类型的值,但是在检测引用类型的值时用处不大。可以使用instanceof。 arr instanceof Array 判断一个对象是不是某个类型(只能判断引用类型) delete: 删除对象属性、数组元素、变量 例:delete a.name;(delete a["name"]) 删除数组中的值:delete arr[0];(arr[0]=undefined) 删除变量:delete a; 来源:

How to get rid of instanceof in this Builder implementation

拟墨画扇 提交于 2019-12-05 11:02:46
The idea I need to create Commands. Commands can be configured with parameters. Not every command can receive the same parameters. So some have to be ignored. I have an abstract class Command in which I've defined a Builder. By default every append parameter throws 'UnsupportedOperationException' public abstract class Command { public static abstract class CommandBuilder { // TODO instanceof. How to do this better? public CommandBuilder append(Parameter p) throws UnsupportedOperationException { if (p instanceof URLParameter) return append((URLParameter) p); if (p instanceof ActionParameter)

instanceof in Java - why doesn't this compile? [duplicate]

浪尽此生 提交于 2019-12-05 10:38:01
问题 This question already has answers here : Why does `instanceof` error rather than return `false` when used for 2 incompatible classes? (4 answers) Closed 6 years ago . class A { public static void main(String...args) { Integer var = 10; if(var instanceof Character) // Line1 System.out.println("var is a Character"); } } I know Line 1 will not compile because the compiler has found that var is not a Character . What I fail to understand is why the compiler throws an error instead of returning

How to “instanceof” a primitive string (string literal) in JavaScript [duplicate]

给你一囗甜甜゛ 提交于 2019-12-05 08:33:52
问题 This question already has answers here : Why does instanceof return false for some literals? (10 answers) Closed 6 years ago . In JavaScript, I can declare a string in the following ways; var a = "Hello World"; var b = new String("Hello World"); but a is not an instance of String... console.log(a instanceof String); //false; console.log(b instanceof String); //true; So how do you find the type or " instanceof " a string literal? Can JavaScript be forced to create a new String() for every