CODEWITHSUNDEEP
java
e2rabi
e2rabi

Reputation: 4848

Confused about comparing string in this example

I know that "==" compare reference and the a java String is immutable and use a string pool cache but I'm still confusing on this example : 

 final String fName = "James";
        String lName = "Gosling";
        String name1 = fName + lName;
        String name2 = fName + "Gosling";
        String name3 = "James" + "Gosling";
       System.out.println(name1 == name2);  // 1
        System.out.println(name2 == name3);

This gave me as results :

false
true

I still confused why System.out.println(name1 == name2); give me a false as I know the both values should be cached in the string pool ?

Upvotes: 0

Views: 138

Answers (2)

JB Nizet
JB Nizet

Reputation: 692231

Since fName is final and initialized with a literal String, it's a constant expression.

So the instruction

String name2 = fName + "Gosling"

is compiled to 

String name2 = "James" + "Gosling"

which is compiled to

String name2 = "JamesGosling"

So, in the bytecode, you have the equivalent of 

String name2 = "JamesGosling";
String name3 = "JamesGosling";

So both name2 and name3 reference the same literal String, which is interned.

On the other hand, lName is not final, and is thus not a constant expression, and the concatenation happens at runtime rather than at compile time. So the concatenation creates a new, non-interned String.

Upvotes: 3

But I'm Not A Wrapper Class
But I'm Not A Wrapper Class

Reputation: 14296

Creating two strings with same string values does not guarantee they are both from the String Pool. You need to call s.intern() to ensure you get a String that has both same value and memory address from String Pool.

There is a good chance the concatenation is causing the String to be read as String Object instead of a String literal. This is due to the compiler not knowing if fName + lName will result to a literal or not. If it's a String literal, it goes to the String Pool. If it's a String Object, it is is not guaranteed to go to the String Pool and be treated like any other Object.

Upvotes: 1

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