Difference in the results of String comparison in Java? [duplicate]

Question

How does Java implement the below string comparisons

public class MyClass { public static void main(String args[]) { String a = "Chaipau"; String b = "pau"; System.out.println(a == "Chai"+"pau"); //true System.out.println(a == "Chai"+b); //false } }

This is different from How do I compare strings in Java? , as the answer does not contain the why a new object is created in the second case , when it could have pointed to the same object reference as it is done in the first case.

Don't compare strings like that. stackoverflow.com/questions/513832/… — Worthless
– Worthless, Commented Sep 11, 2018 at 11:18
that's a referential comparison, don't expect it to behave like a comparison of values — Stultuske
– Stultuske, Commented Sep 11, 2018 at 11:19
The == compares references, so the "numbers" that identify the two memory areas occupied by the objects. It is a 32 or 64 bit value depending on platform. In conclusion, the result is non deterministic, because it depends on memory actual state; that's the reason because == should not be used when comparing non-primitive types. — Andrea
– Andrea, Commented Sep 11, 2018 at 11:24
the first one is a constant and it is interned in the string constant pool, thus two references point to the same thing in the constant pool. the second one is not a constant, but you try to reference it from the constant pool, thus false — Eugene
– Eugene, Commented Sep 11, 2018 at 12:01

Andy Turner · Accepted Answer · 2018-09-11 11:24:45Z

5

"Chai"+"pau" is semantically identical to "Chaipau", and thus is the same instance that a refers to.

"Chai"+b is evaluated at runtime (because b is not a compile-time constant expression), creating a new instance of String, and thus is not the same instance that a refers to.

answered Sep 11, 2018 at 11:24

Andy Turner

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2 Comments

Michael Over a year ago

Specifically, "Chai"+"pau" is a compile-time constant: "Constant expressions of type String are always "interned" so as to share unique instances"

Chaipau Over a year ago

Cool, as declaring final String b = "pau"; leaves the variable b to a compile time expression and the second case results to true.

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