String.intern() 显示奇怪的结果，它是如何工作的

Question

I know that String.intern() adds string to pool if it does not contain the object, but how to explain the results.

The code below:

public static void main(String[] args) {

        char[] abc = new char[]{'a','b','c'};
        String str = new String(abc);
        System.out.println(str == "abc");
        str.intern();
        System.out.println(str == "abc");
    }

output is:

false

false

However when code as below:

public static void main(String[] args) {

        char[] abc = new char[]{'a','b','c'};
        String str = new String(abc);
        str.intern();
        System.out.println(str == "abc");
    }

The output is:

true

What's the difference.

Answer 1

The difference is that when you use the String literal "abc" before the explicit call to intern , it is implicitly interned.

str.intern() doesn't store the instance referenced by str in the String pool if an equal String is already in the pool.

First snippet:

System.out.println(str == "abc"); // "abc" is interned
str.intern(); // str is not stored in the pool
System.out.println(str == "abc"); // returns false

Second snippet:

str.intern(); // str is stored in the pool
System.out.println(str == "abc"); // returns true

From the Javadoc of intern:

String java.lang.String.intern()

When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the equals(Object) method, then the string from the pool is returned . Otherwise, this String object is added to the pool and a reference to this String object is returned... All literal strings and string-valued constant expressions are interned .