Detailed explanation of string constant pool and intern() of string

String constant pool

One 、 What is a constant pool ?

Constant pool stay java It is used to save the... Determined at compile time , Compiled class A piece of data in the file , Includes about classes , Method , Constants in interfaces, etc , It also includes string constants .
String constant pool It's a fixed size Hashtable, Used to store strings .
In the pile (heap) A part of memory space is allocated as constant pool area .

Two 、 Why should there be a string constant pool ?

String Class is used very frequently .
for fear of JVM First, create a large number of string objects , Then garbage collection .
To save memory and improve efficiency .

3、 ... and 、 How to generate objects in the string constant pool ?

1. The main generation methods

1. Use it directly Double quotes It was announced String object , direct Stored in constant pool in .
2. Use it directly new keyword The created string object will First, look in the string constant pool , If not found, create one in the string constant pool , Then create a string object in the heap ; If you find it , Just create a string object directly in the heap .
3. If Not in double quotes Declarative String object , Not directly new Out object , have access to intern Method realization Add constant pool .（ Such as String s1 = new String(“hello”) + new String(“world”); Created s1 object ）

2. String Splicing generation

1 String splicing

public class StringTest {
    
    public static void main(String[] args) {
    
        String a = "wechat" + "zhou";
    }
}

wechat and zhou All strings , At compile time , The compiler will automatically optimize this line of code For the following code

public class StringTest {
    
    public static void main(String[] args) {
    
      String a = "wechatzhou";  
    }
}

2 Reference splicing

（ In fact, it is the third point of the above main generation methods , Need to use intern（） Method ）
Reference splicing ( Such as s3) The final content of , Will not exist in the constant pool
String s1=“1”; Created 1 individual "1" object , In the constant pool .
String s2=new String(“1”); I created two "1" object , One in constant pool , In a heap District .
String s3=“1”+“1”; Created a "11" object , Not in constant pool , stay heap District .

public class StringTest {
    
    public static void main(String[] args) {
    
        String a = "we";
      	String b = a + "chat";
      	//  The upper and lower meanings are the same 
      	String c = "moz";
      	String d = "bad";
      	String f = c + d;
    }
}

Because the referenced value cannot be determined at compile time
When there is a string reference （a+"chat" or c+d） When splicing ,Java The compiler will create a StringBuilder object , adopt append() Method to realize splicing .
"wechat" It will not be added to the string constant pool , however Can pass b.intern() Method is added to the constant pool

3 final Splicing

The effect is consistent with string splicing

public class StringTest {
    
    public static void main(String[] args) {
    
        final String a = "we";
     		final String b = "chat";
      	String c = a + b + "handsome";
    }
}

use final The modified string is known at compile time Of , The above code will be optimized to

public class StringTest {
    
    public static void main(String[] args) {
    
       			final String a = "we";
     			final String b = "chat";
				String c= "wechathandsome";
    }
}

Intern（） Method

One 、intern（） The role of methods

You can add to the constant pool
1 If in the constant pool There is The current string , will Go straight back to The current string .
2 If in the constant pool non-existent The current string , This string Put into constant pool , Back again .

Two 、intern（） Return value of method

1 If in the constant pool There is The current string , will Go straight back to Reference to the current string .
2 If in the constant pool non-existent The current string ：
if heap The object of this string exists in the area , Will The object is moved to the constant pool , And return the reference of the object .
if heap The object does not exist in the zone , It's in Constant pool creates this string , And return a reference to the string .
（String s="1" Equivalent to String s=“1”.intern()）

Do the title

The constant pool is in the heap , But for ease of description , The following heap areas refer to the places after the heap area is removed from the constant pool area .

One 、String s = new String(“abc”) Several objects are created ？

    Created 1 or 2 Objects .（ Depends on whether the string constant pool originally exists abc character string ）
1 If the string constant pool already exists abc character string , Then create an object , In the pile area .
2 If the string constant pool does not exist abc character string , Then create two objects , One in constant pool , One in the pile area .

Two 、 The output of the following two pieces of code

jdk6 and jdk The difference between districts ： The constant area moves from outside the heap area to inside the heap area .

code snippet 1

public static void main(String[] args) {
    
    String s = new String("1");
    s.intern();
    String s2 = "1";
    System.out.println(s == s2);

    String s3 = new String("1") + new String("1");
    s3.intern();
    String s4 = "11";
    System.out.println(s3 == s4);
}

jdk6 Next false false
jdk7 Next false true

about s3 and s4

String s3 = new String(“1”) + new String(“1”);
Generated 2 End object , Namely In the string constant pool “1” and JAVA Heap Medium s3 The object to which the reference refers . In the middle 2 Anonymous new String(“1”) No discussion .
here s3 The reference object is ”11”, There is no “11”.
s3.intern(); take s3 Medium “11” String into String Constant pool in .
jdk6： stay perm Generate a... In the string constant pool of the area “11” The object of .
jdk7： Constant pool is not Perm Area and in heap District , There is no need to create another object , It is You can store references in the heap directly . This quote points to s3 Referenced object . That is to say, the reference address is the same .
Last String s4 = “11”; In this code ”11” It's a statement , It will be directly created in the constant pool , It is found that this string already exists , That is to point s3 A reference to an object . therefore s4 The quotation points to and s3 The same . So the final comparison s3 == s4 yes true.

about s and s2

String s = new String(“1”); Generated 2 Objects . In the constant pool “1” and JAVA Heap String object in .
s.intern(); Go to the constant pool and find “1” It's already in the constant pool , Do not change .
String s2 = “1”; Generate a s2 References to point to... In the constant pool “1” object .
and s Point to JAVA Heap String object in , Does not point to constant pool , Therefore, the two are not equal .

code snippet 2

public static void main(String[] args) {
    
    String s = new String("1");
    String s2 = "1";
    s.intern();
    System.out.println(s == s2);

    String s3 = new String("1") + new String("1");
    String s4 = "11";
    s3.intern();
    System.out.println(s3 == s4);
}

jdk6 Next false false
jdk7 Next false false

about s3 and s4

String s3 = new String(“1”) + new String(“1”); Generated 2 End object , Namely In the string constant pool “1” and JAVA Heap Medium s3 The object to which the reference refers .
String s4 = “11”; There is no constant pool “11” object , Create... In the string constant pool “11” String .s4 Point there .
s3.intern();, In constant pool “11” The object already exists , So there is no change .s3 Still original heap An object of the zone （ Not in constant pool ）.s3 And s4 Different .

about s and s2

String s = new String(“1”); Has been generated in the constant pool “1” Object .
s.intern(); No change .
s2 Declarations are all address references directly from the constant pool .
s Point to heap District （ Outside the constant pool ） The object of “1”.
s and s2 The reference addresses of are not equal .

Reference link 1
Reference link 2
See the link 3