CODEWITHSUNDEEP
javacomparison
Manu
Manu

Reputation: 3247

java Long datatype comparison

Why does the code below return false for long3 == long2 comparison even though it's literal.

public class Strings {

    public static void main(String[] args) {
        Long long1 = 256L + 256L;
        Long long2 = 512L;
        Long long3 = 512L;
        System.out.println(long3 == long2);
        System.out.println(long1.equals(long2));
    }
}

Upvotes: 33

Views: 56022

Answers (4)

TheSprinter
TheSprinter

Reputation: 1528

Here Long is a Wrapper class so the below line will compare the reference not the content.

long3 == long2

its always better to compare with ** .longValue() ** like below

long3.longValue() == long2.longValue()

If we use in-build equal() method that also will do the same thing with null check.

long3.equals(long2)

Below is the internal implementation of equals() in java

public boolean equals(Object obj) {
    if (obj instanceof Long) {
        return value == ((Long)obj).longValue();
    }
    return false;
}

Upvotes: 1

madhu_karnati
madhu_karnati

Reputation: 795

If you want to do 

      str3==str2

do like this.. 

     str3.longValue()==str2.longValue()

This serves your purpose and much faster because you are comparing two primitive type values not objects.

Upvotes: 3

Brian Roach
Brian Roach

Reputation: 76908

Long is an object, not a primitive. By using == you're comparing the reference values.

You need to do:

if(str.equals(str2))

As you do in your second comparison.

Edit: I get it ... you are thinking that other objects act like String literals. They don't*. And even then, you never want to use == with String literals either.

(*Autobox types do implement the flyweight pattern, but only for values -128 -> 127. If you made your Long equal to 50 you would indeed have two references to the same flyweight object. And again, never use == to compare them. )

Edit to add: This is specifically stated in the Java Language Specification, Section 5.1.7:

If the value p being boxed is true, false, a byte, or a char in the range \u0000 to \u007f, or an int or short number between -128 and 127 (inclusive), then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.

Note that long is not specifically mentioned but the current Oracle and OpenJDK implementations do so (1.6 and 1.7), which is yet another reason to never use ==

Long l = 5L;
Long l2 = 5L;
System.out.println(l == l2);
l = 5000L;
l2 = 5000L;
System.out.println(l == l2);

Outputs:

true
false

Upvotes: 66

Mr_CRivera
Mr_CRivera

Reputation: 239

You could also get the primitive value out of the Long object using:

str.longValue()

Upvotes: 3

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