This question already has an answer here: Is Java “pass-by-reference” or “pass-by-value”? 78 answers Java uses pass by value, not by reference... But, for non primitive types the value is the value of the reference. So == compares the values of references for Objects. For a detailed explanation, see my article "Java is Pass-By-Value, Dammit!" http://javadude.com/articles/pa
这个问题在这里已经有了答案： Java是“通过引用传递”还是“按值传递”？ 78个答案 Java使用按值传递，而不是引用... 但是，对于非基元类型，值是引用的值。 所以==比较对象的引用值。 有关详细的解释，请参阅我的文章“Java是按值传递，该死！” http://javadude.com/articles/passbyvalue.htm 区别之处在于“通过引用 ** ”和“传递 **引用”之间。 您有时也会看到“通过......呼叫”和“通过......”交替使用。 为了简单
This question already has an answer here: Is Java “pass-by-reference” or “pass-by-value”? 78 answers In the case of Arraylist string objects the added elements are getting retrived. In case of String the method call has no effect on the String being passed. It happens cause Java is Pass-by-Value and String s are immutable When you call markAsNull(ArrayList<String> str) The a new
这个问题在这里已经有了答案： Java是“通过引用传递”还是“按值传递”？ 78个答案 在Arraylist字符串对象的情况下，添加的元素正在回归。 在String的情况下，方法调用对传递的字符串没有影响。 这是因为Java是Pass-by-Value和String是不可变的 你打电话时 markAsNull(ArrayList<String> str) 通过名称str的新引用为al指向的同一个ArrayList创建。 当你在str上add一个元素时，它会被添加到同一个对象中。 稍后
This question already has an answer here: Is Java “pass-by-reference” or “pass-by-value”? 78 answers The array is passed by reference, but the reference is passed by value. That is, you can change the array that a refers to, but you cannot change which array a refers to. Java is pass by value. This is why your code does not work. A good practice would be to mark int[] a as final so this
这个问题在这里已经有了答案： Java是“通过引用传递”还是“按值传递”？ 78个答案 该数组通过引用传递，但引用是按值传递的。 也就是说，你可以改变阵列a指，但你不能改变其阵列a指。 Java是通过价值传递的。 这就是为什么你的代码不起作用。 一个好的做法是将int[] a标记为final ，这样会导致编译错误（请参阅相应的Checkstyle规则）。 从函数返回参数“a”并将其分配给主函数中的testArray。 当您通过引用传递对象时
Possible Duplicate: Is Java pass by reference? public class myClass{ public static void main(String[] args){ myObject obj = new myObject("myName"); changeName(obj); System.out.print(obj.getName()); // This prints "anotherName" } public static void changeName(myObject obj){ obj.setName("anotherName"); } } I know that Java pass by value, but why doe
可能重复：是否通过引用传递Java？ public class myClass{ public static void main(String[] args){ myObject obj = new myObject("myName"); changeName(obj); System.out.print(obj.getName()); // This prints "anotherName" } public static void changeName(myObject obj){ obj.setName("anotherName"); } } 我知道Java按值传递，但为什么它在前面的示例中通过引用传递
Possible Duplicate: Is Java “pass-by-reference”? I found an unusual Java method today: private void addShortenedName(ArrayList<String> voiceSetList, String vsName) { if (null == vsName) vsName = ""; else vsName = vsName.trim(); String shortenedVoiceSetName = vsName.substring(0, Math.min(8, vsName.length())); //SCR10638 - Prevent export of empty rows.
可能重复： Java是否“通过引用”？ 我今天发现了一种不寻常的Java方法： private void addShortenedName(ArrayList<String> voiceSetList, String vsName) { if (null == vsName) vsName = ""; else vsName = vsName.trim(); String shortenedVoiceSetName = vsName.substring(0, Math.min(8, vsName.length())); //SCR10638 - Prevent export of empty rows. if (shortenedVoi
Possible Duplicate: Is Java “pass-by-reference”? Arrays are not a primitive type in Java, but they are not objects either, so are they passed by value or by reference? Does it depend on what the array contains, for example references or a primitive type? Your question is based on a false premise. Arrays are not a primitive type in Java, but they are not objects either ... " In fac
可能重复： Java是否“通过引用”？ 数组在Java中不是原始类型，但它们也不是对象，它们是按值还是按引用传递的？ 它取决于数组包含的内容，例如引用或基本类型？ 你的问题是基于一个错误的前提。 数组在Java中不是原始类型，但它们不是对象......“ 事实上，Java 中的所有数组都是 objects1。 每个Java数组类型都有java.lang.Object作为其超类型，并继承Object API中所有方法的实现。 像所有Java对象一样，数组通过
Can we print something without using Main(public static void main string[]args) method in java? i have tried using static block it's not working in java 8 version.just curious You need to somehow indicate a point where execution of the programs starts. This is called "an entry point". In some languages you start execution from the first line of code. Perl is an example. In Ja
我们可以打印一些东西，而不使用java中的Main（public static void main string [] args）方法吗？ 我曾尝试使用静态块它不工作在Java 8 version.just好奇 你需要以某种方式指出程序执行的开始点。 这被称为“入口点”。 在某些语言中，您从第一行代码开始执行。 Perl是一个例子。 在Java中，你从一个标记为public static void main(String[] args){/* do something */} aka psvm 。 这使您可以灵活地根据程序逻辑安排代码
This question already has an answer here: Are fields initialized before constructor code is run in Java? 4 answers Instance variable initialization expressions such as int var = getVal(); are evaluated prior to the constructor execution. Therefore getVal() is called before the constructor is executed. Constructor is called prior to method. The execution of method occurs after that which
这个问题在这里已经有了答案： 在Java中运行构造函数代码之前，字段是否已初始化？ 4个答案 实例变量初始化表达式，如int var = getVal(); 在构造函数执行之前进行评估。 因此，在执行构造函数之前调用getVal() 。 构造函数在方法之前调用。 方法的执行发生在作为对象实例变量评估对象创建的一部分之后。 从以下代码可以更好地理解这一点。 class SuperClass{ SuperClass(){ System.out.println("Super c
The following print statement would print "hello world". Could anyone explain this? System.out.println(randomString(-229985452) + " " + randomString(-147909649)); And randomString() looks like this: public static String randomString(int i) { Random ran = new Random(i); StringBuilder sb = new StringBuilder(); while (true) { int k = ran.nextInt(27); if
以下打印声明将打印“hello world”。 有谁能解释这个吗？ System.out.println(randomString(-229985452) + " " + randomString(-147909649)); 和randomString()看起来像这样： public static String randomString(int i) { Random ran = new Random(i); StringBuilder sb = new StringBuilder(); while (true) { int k = ran.nextInt(27); if (k == 0) break; sb.append
public class Main { public static void main(String[] args) { System.out.println(B.x); } } class A { public static String x = "x"; } class B extends A { static { System.out.print("Inside B."); } } Question: Why output will be: "x". But not: "Inside Bx" The reference to Bx issues the fo
public class Main { public static void main(String[] args) { System.out.println(B.x); } } class A { public static String x = "x"; } class B extends A { static { System.out.print("Inside B."); } } 问题：为什么输出将是：“x”。 但不是：“内部Bx” 对Bx的引用发出以下字节码： getstatic #3 <Field int B