Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
427 views
in Technique[技术] by (71.8m points)

java - The return of String.intern() explained

Consider:

String s1 = new StringBuilder("Cattie").append(" & Doggie").toString();
System.out.println(s1.intern() == s1); // true why?
System.out.println(s1 == "Cattie & Doggie"); // true another why?

String s2 = new StringBuilder("ja").append("va").toString();
System.out.println(s2.intern() == s2); // false

String s3 = new String("Cattie & Doggie");
System.out.println(s3.intern() == s3); // false
System.out.println(s3 == "Cattie & Doggie"); // false

I got confused why they are resulting differently by the returned value of String.intern() which says:

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.

Especially after these two tests:

assertFalse("new String() should create a new instance", new String("jav") == "jav");
assertFalse("new StringBuilder() should create a new instance",
    new StringBuilder("jav").toString() == "jav");

I once read a post talking about some special strings interned before everything else, but it's a real blur now.

If there are some strings pre-interned, is there a way to get kind of a list of them? I am just curious about what they can be.


Updated

Thanks to the help of @Eran and @Slaw, I finally can explain what just happened there for the output

true
true
false
false
false
  1. Since "Cattie & Doggie" doesn't exist in the pool, s1.intern() will put the current object reference to the pool and return itself, so s1.intern() == s1;
  2. "Cattie & Doggie" already in the pool now, so string literal "Cattie & Doggie" will just use the reference in pool which is actually s1, so again we have true;
  3. new StringBuilder().toString() will create a new instance while "java" is already in the pool and then the reference in pool will be returned when calling s2.intern(), so s2.intern() != s2 and we have false;
  4. new String() will also return a new instance, but when we try to s3.intern(), it will return the previously stored reference in the pool which is actualy s1 so s3.intern() != s3 and we have false;
  5. As #2 already discussed, String literal "Cattie & Doggie" will return the reference already stored in the pool (which is actually s1), so s3 != "Cattie & Doggie" and we have false again.

Thanks for @Sunny to provide a trick to get all the interned strings.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

s2.intern() would return the instance referenced by s2 only if the String pool didn't contain a String whose value is "java" prior to that call. The JDK classes intern some Strings before your code is executed. "java" must be one of them. Therefore, s2.intern() returns the previously interned instance instead of s2.

On the other hand, the JDK classes did not intern any String whose value is equal to "Cattie & Doggie", so s1.intern() returns s1.

I am not aware of any list of pre-interned Strings. Such a list will most likely be considered an implementation detail, which may vary on different JDK implementations and JDK versions, and should not be relied on.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...