抱歉回复晚了，但我急需回复。 首先，我们需要知道一些Java.lang.String类规则。

String Literals e.g.String str="java"; (we use only double Quotes) are different from String Object (we use new keyword) e.g. String str=new String("java"); String is Immutable Object i.e. If value changes a new Object is created and returned to you eg See replace() and replaceAll() functions and many more. This creates a problem of many String Object in Modification, So creators of Java came up an Idea was called StringPool. StringPool is stored in heap area where object reference data will be stored as we know String is Char[](before java 9 very Long to read) or byte[](after java 9 short to read). String literals are stored in StringPool and String Objects are stored in as usual heap Object Area. If there are many Object String Initialization JVM heap will be finished in String Operations only, Java Development team came up with intern() solution this moves/changes memory reference to StringPool. Program: Comparing String references to objects

另一个更好地理解java.lang.String的好链接

import java.util.*; 

class GFG { 
    public static void main(String[] args) 
    { 
      String siteName1 = "java.com";
        String siteName2 = "java.com";
        String siteName3 = new String("java.com");
        String siteName4 = new String("java.com").intern();
      
    System.out.println("siteName1:::"+Integer.toHexString(System.identityHashCode(siteName1)));
      System.out.println("siteName2:::"+Integer.toHexString(System.identityHashCode(siteName2)));
      System.out.println("siteName3 creation Of New Object Without Interned:::"+Integer.toHexString(System.identityHashCode(siteName3)));//must be Diffrent bcoz new Object In Heap Area
      System.out.println("siteName4 creation Of New Object With Interned:::"+Integer.toHexString(System.identityHashCode(siteName4)));//must be same MemoryAddress of siteName1,siteName2 and Interned, bcoz Objects Points to String pool Now
      
      System.out.println(siteName1 == siteName2); // true
      System.out.println(siteName1 == siteName3); // false this tells about lietral vs String Objects
      String siteName5 = siteName3.intern(); // Interning will not change Original Object but gives us a new Object
      
      System.out.println("siteName5 Interned from siteName3:::"+Integer.toHexString(System.identityHashCode(siteName5)));//must be same MemoryAddress of siteName1,siteName2 and Interned, bcoz Objects Points to String pool Now
      
      System.out.println(siteName1 == siteName3); // false this tells about Immutability
      System.out.println(siteName1 == siteName5); // true After Intering both are same
      System.out.println(siteName1 == siteName4); // true
      System.out.println(siteName5 == siteName4); // true
    } 
}

当您将字符串存储为

String string1 = "Hello";

然后JVM在称为String常量池的单独内存块中创建一个具有给定价格的String对象。

当我们有尝试生成另一个String的倾向时

String string2 = "Hello";

JVM验证String常量池中是否存在价格恒定的String对象，如果存在，JVM不会创建一个全新的对象，而是将现有对象的引用赋值给新变量。

当我们存储String为

String string = new String("Hello");

使用new关键字，不管String常量池的内容是什么，都会生成一个具有给定价格的全新对象。

把“bla”想象成一个魔法工厂，就像Strings.createString(“bla”)(pseudo)。工厂保存了所有以这种方式创建的字符串池。

如果它被调用，它会检查池中是否已经有这个值的字符串。如果为真，则返回此string对象，因此以这种方式获得的字符串确实是同一个对象。

如果不是，它在内部创建一个新的字符串对象，将其保存在池中，然后返回它。因此，当下次查询相同的字符串值时，它将返回相同的实例。

手动创建新的String("")通过绕过字符串字面量池来覆盖这种行为。因此，应该始终使用equals()检查相等性，它比较字符序列，而不是对象引用相等性。

抱歉回复晚了，但我急需回复。 首先，我们需要知道一些Java.lang.String类规则。

String Literals e.g.String str="java"; (we use only double Quotes) are different from String Object (we use new keyword) e.g. String str=new String("java"); String is Immutable Object i.e. If value changes a new Object is created and returned to you eg See replace() and replaceAll() functions and many more. This creates a problem of many String Object in Modification, So creators of Java came up an Idea was called StringPool. StringPool is stored in heap area where object reference data will be stored as we know String is Char[](before java 9 very Long to read) or byte[](after java 9 short to read). String literals are stored in StringPool and String Objects are stored in as usual heap Object Area. If there are many Object String Initialization JVM heap will be finished in String Operations only, Java Development team came up with intern() solution this moves/changes memory reference to StringPool. Program: Comparing String references to objects

另一个更好地理解java.lang.String的好链接

import java.util.*; 

class GFG { 
    public static void main(String[] args) 
    { 
      String siteName1 = "java.com";
        String siteName2 = "java.com";
        String siteName3 = new String("java.com");
        String siteName4 = new String("java.com").intern();
      
    System.out.println("siteName1:::"+Integer.toHexString(System.identityHashCode(siteName1)));
      System.out.println("siteName2:::"+Integer.toHexString(System.identityHashCode(siteName2)));
      System.out.println("siteName3 creation Of New Object Without Interned:::"+Integer.toHexString(System.identityHashCode(siteName3)));//must be Diffrent bcoz new Object In Heap Area
      System.out.println("siteName4 creation Of New Object With Interned:::"+Integer.toHexString(System.identityHashCode(siteName4)));//must be same MemoryAddress of siteName1,siteName2 and Interned, bcoz Objects Points to String pool Now
      
      System.out.println(siteName1 == siteName2); // true
      System.out.println(siteName1 == siteName3); // false this tells about lietral vs String Objects
      String siteName5 = siteName3.intern(); // Interning will not change Original Object but gives us a new Object
      
      System.out.println("siteName5 Interned from siteName3:::"+Integer.toHexString(System.identityHashCode(siteName5)));//must be same MemoryAddress of siteName1,siteName2 and Interned, bcoz Objects Points to String pool Now
      
      System.out.println(siteName1 == siteName3); // false this tells about Immutability
      System.out.println(siteName1 == siteName5); // true After Intering both are same
      System.out.println(siteName1 == siteName4); // true
      System.out.println(siteName5 == siteName4); // true
    } 
}

String str = new String("hello")

它将检查字符串常量池是否已经包含字符串“hello”? 如果存在，则不会在String常量池中添加条目。如果不存在，它将在String常量池中添加一个条目。

对象将在堆内存区域中创建，str引用指向在堆内存位置中创建的对象。

如果你想要str引用String常量池中包含的点对象，则必须显式调用str.intern();

String str = "world";

它将检查字符串常量池是否已经包含字符串“hello”? 如果存在，则不会在String常量池中添加条目。如果不存在，它将在String常量池中添加一个条目。

在上述两种情况下，str引用指向常量池中存在的字符串“world”。

新的字符串(“文本”); 显式地创建String对象的一个新的引用不同的实例;字符串s = "text";可以重用字符串常量池中的实例(如果有)。

你很少会想要使用新的String(anotherString)构造函数。来自API:

String(String original):初始化新创建的String对象，使其表示与参数相同的字符序列;换句话说，新创建的字符串是参数字符串的副本。除非需要original的显式副本，否则使用此构造函数是不必要的，因为字符串是不可变的。

相关问题

Java字符串:“字符串s =新字符串(“愚蠢的”);” 字符串是Java中的对象，所以为什么我们不使用' new '来创建它们呢?

什么是指涉区别

检查下面的代码片段:

    String s1 = "foobar";
    String s2 = "foobar";

    System.out.println(s1 == s2);      // true

    s2 = new String("foobar");
    System.out.println(s1 == s2);      // false
    System.out.println(s1.equals(s2)); // true

两个引用类型上的==是引用标识符比较。两个相等的对象不一定是==。在引用类型上使用==通常是错误的;大多数时候应该用等号代替。

尽管如此，如果出于某种原因需要创建两个equals而不是== string，则可以使用新的string (anotherString)构造函数。然而，需要再次说明的是，这是非常奇怪的，而且很少是故意的。

参考文献

JLS 15.21.3引用相等操作符==和!= 类对象-布尔对象(等于)

相关问题

Java字符串。等于和== 我如何比较字符串在Java?