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In evaluating the expression (x==y && a<b) the boolean expression x==y is evaluated first and then a<b is evaluated in Java ? (True or False):

I find False as the answer but I can't reason it. Moreover, according to my C/C++ knowledge, if x==y evaluated to false, the whole expression (x==y && a<b) evaluates to false without even moving on to evaluate a<b.

If this was similar for Java, the answer to the statement must have been True.

I couldn't find any suitable explanation for the statement to be False, so please help me out.

asked May 31 '19 at 04:52

Biswajit Roy

[Java Operator Precedence](https://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html) – Tibrogargan May 31 '19 at 04:55
This link explains it well - https://stackoverflow.com/questions/8759868/java-logical-operator-short-circuiting – Anshul Singhal May 31 '19 at 04:57
Just, a clarification, after reading I feel that answer is `True`. Is that correct and the book somehow made a misprint ? – Biswajit Roy May 31 '19 at 05:02

Java: boolean expression (x==y && a

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