Understanding effectively final variable in Java

[Swapna latha]

Hello everyone, I have tried and executed a program using method references but got confused in understanding effectively final variable in java in the following code :

package lambdas.demo17;

import java.util.function.Predicate;

public class PredicateMethodReferenceTest {
    public static void main(String[] args) {
        String str ="ABC";// String reference i.e., Object reference, Here str is a effectively final variable
        str = "XYZ";// reassigning the effectively final variable, so lambda or method reference should not work, but it worked here , Its really strange. 
        Predicate<String> stringPredicate = str::contains;
        boolean flag = stringPredicate.test("S");
        System.out.println(flag);
        str = "SURYA";
        Predicate<String> stringPredicate2 = str::contains;
        System.out.println(flag);
    }
}

If str is not effectively final, how does the program compiled and ran with output.

[Tim Cooke]

What makes you think line 7 String str ="ABC"; is an "effectively final" variable?

[Mikalai Zaikin]

Your code is not a lambda expression but rather a method reference. And they have indeed different restrictions.

While lambda cannot use non-final local variables, method reference has no such limitation.

Please check the details here: https://stackoverflow.com/questions/33052917/why-can-method-reference-use-non-final-variables

[Swapna latha]

Thank you so much for posting the stackoverflow link. I understood now.

Mikalai Zaikin wrote:Your code is not a lambda expression but rather a method reference. And they have indeed different restrictions.

While lambda cannot use non-final local variables, method reference has no such limitation.

Please check the details here: https://stackoverflow.com/questions/33052917/why-can-method-reference-use-non-final-variables

[Mike Simmons]

Mikalai is quite correct that the central issue in your question is answered by the fact that you are not, in fact, using lambdas here. You are using method references, and they have different rules.  They have no problem with variables being final or effectively final.

The stack overflow link refers in several places to "final variables" when it should really talk about effectively final variables.  So be careful when reading it; some of the statements may be misleading.

Your code example does not have any effectively final variables.  The variable on line 7 could have been effectively final, if you had not added code on lines 8 and 12 which changes it.  Once you do that, the variable is no effectively final anywhere, even at line 7.