一个是在字符串常量池中创建一个字符串

String s = "text";

另一个在常量池中创建一个字符串("text")，在普通堆空间中创建另一个字符串(s)。两个字符串的值相同，即"text"。

String s = new String("text");

如果以后未使用，则会丢失(符合GC的条件)。

另一方面，字符串字面值被重用。如果你在你的类的多个地方使用“text”，它实际上将是一个且只有一个String(即在池中多次引用同一个字符串)。

@Braj:我想你已经提到了另一种方式。如果我错了，请指正

逐行创建对象:

字符串 str1 = new String（“java5”）

   Pool- "java5" (1 Object)

   Heap - str1 => "java5" (1 Object)

String str2 = "java5"

  pool- str2 => "java5" (1 Object)

  heap - str1 => "java5" (1 Object)

字符串 str3 = 新字符串（str2）

  pool- str2 => "java5" (1 Object)

  heap- str1 => "java5", str3 => "java5" (2 Objects)

String str4 = "java5"

  pool - str2 => str4 => "java5" (1 Object)

  heap - str1 => "java5", str3 => "java5" (2 Objects)

字符串字面值将进入字符串常量池。

下面的快照可能会帮助你从视觉上理解它，从而更长时间地记住它。

逐行创建对象:

String str1 = new String("java5");

在构造函数中使用字符串字面值“java5”，新的字符串值存储在字符串常量池中。 使用new操作符，在堆中创建一个以“java5”为值的新字符串对象。

String str2 = "java5"

引用“str2”指向字符串常量池中已经存储的值

String str3 = new String(str2);

在堆中创建一个新的字符串对象，其值与"str2"引用的值相同。

String str4 = "java5";

引用“str4”指向字符串常量池中已经存储的值

对象总数:堆- 2，池- 1

Oracle社区的进一步阅读

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
    } 
}

尽管从程序员的角度来看，它们看起来是一样的，但它对性能有很大的影响。你几乎总是想用第一种形式。

一个简单的理解差异的方法如下:-

String s ="abc";
String s1= "abc";
String s2=new String("abc");

        if(s==s1){
            System.out.println("s==s1 is true");
        }else{
            System.out.println("s==s1 is false");
        }
        if(s==s2){
            System.out.println("s==s2 is true");
        }else{
            System.out.println("s==s2 is false");
        }

输出是

s==s1 is true
s==s2 is false

因此new String()总是会创建一个新实例。