Java String literals concatenation

Question

    public static void main(String[] args){
        one();
        two();
        three();
    }

    public static void one() {
        String s1 = "hill5";
        String s2 = "hill" + 5;
        System.out.println(s1==s2);
    }

    public static void two() {
        String s1 = "hill5";
        int i =5;
        String s2 = "hill" + i;
        System.out.println(s1==s2);
    }

    public static void three() {
        String s1 = "hill5";
        String s2 = "hill" + s1.length();
        System.out.println(s1==s2);
    }

Output is

true 
false 
false

String literals use interning process,then why two() and three() is false.I can understand in case of three() but two() is not clear.but need proper explanation for both cases.

Can someone please explain proper reason?

read this question which i asked stackoverflow.com/questions/16729045/… — anshulkatta
– anshulkatta, Commented May 27, 2013 at 10:29
For fun, you can also try two with final int i = 5; instead (it will print true instead of false, because now i is a constant). — assylias
– assylias, Commented May 27, 2013 at 10:35
"every time someone compares Strings with == a developer cries..." — Marco Forberg
– Marco Forberg, Commented May 27, 2013 at 10:36
Note that you should not ever rely on == to compare strings - it will make your program brittle. This is simply a question of when strings are interned and that depends on the compiler used (in one()) and the JVM used. — Thorbjørn Ravn Andersen
– Thorbjørn Ravn Andersen, Commented May 27, 2013 at 12:21

Community · Accepted Answer · 2017-05-23 12:33:10Z

14

In case of 2 and 3 , Compiler cannot calculate the value of String , since hill + i is a runtime statement , same for s1.length()

read here which i asked the same case - link

Think like this the String s1 and s2 are using compile time constant , s1="hill5" and s2="hill" + 5 , remember , string assigned as a literal is constant , its state cannot be modified , as String are immutable.

So at Compile time , compiler says "oh yeah , they are calculated as same value , i must assign the same reference to s1 and s2".

But in case of method two() and three() , compiler says " i dont know ,may be value of i can be changed any time , or s1.length() changes any time " , its a runtime thing , so compiler doesn't put s2 of two() and three() method in pool ,

Hence , they are false because at runtime , new object is created as soon it get changed right !!

edited May 23, 2017 at 12:33

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answered May 27, 2013 at 10:31

anshulkatta

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8 Comments

ankita gahoi Over a year ago

But String literals are created as part of heap and jvm check whether String exists or not,if not then it creates new String otherwise points to same location..so in case of two and three hill5 String is there in pool

assylias Over a year ago

@ankitagahoi Not really - only constant strings are put in the pool, unless you use the intern method...

ankita gahoi Over a year ago

all literals not goes to pool?

assylias Over a year ago

@ankitagahoi "hill" + i is not a String literal. "hill5" is... See: docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.5

user207421 Over a year ago

@anshulkatta The JVM doesn't check any such thing. The compiler puts string literals in the constant pool: the OP's code creates new strings. You're just making it up, even after the same statement has already been corrected. Twice. The normative reference is the JLS and the JVM specification, not another forum.

|

Abu Yousuf · Accepted Answer · 2018-07-04 09:37:23Z

1

String with compile time constant expression will be put on String pool. The main condition is compile time constant expression. If you make local variable final in method two() then two() will also print true

public static void two() {
    String s1 = "hill5";
    final int i =5;
    String s2 = "hill" + i;
    System.out.println(s1==s2);
}

Output:

true

answered Jul 4, 2018 at 9:37

Abu Yousuf

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