confused with return value of assignment operation in java

Question

I tried to understand how exactly the return value of assignment operation works. followed by this post "Java returns the assigned value".

    boolean b1 = false, b2 = false;
    if (b2 = b1 == false) {
        System.out.println("true");
    } else {
        System.out.println("false");
    }

b2 is true because of (b1 == false) return true, then the return of b2 assignment b2 = true

Or is it because of some other reason?

its the game of `precedence` of `assignment` and `equal to`. @Pau — jack jay, Jan 07 '17 at 16:34
See here: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html — GhostCat, Jan 07 '17 at 16:42

score 4 · Accepted Answer · answered Jan 07 '17 at 16:35

4

You've got it right. The operator precedence rules make sure that first the == operator is evaluated. That's b1==false, yielding true. After that, the assigned is executed, setting b2 to true. Finally, the assignment operator returns the value as b2, which is evaluated by the if statement.

answered Jan 07 '17 at 16:35

Stephan Rauh

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`int i = doIt() / (j = 2);` , why the method execution comes before the grouping & accessing operators ? they are in the top level of [Operator Precedence and Parentheses](http://stackoverflow.com/documentation/java/176/operators/9207/operator-precedence-and-parentheses#t=201701111907181800707) .thanks – Pauwelyn Jan 11 '17 at 19:09
Java usually evaluates the terms from the left to the right (pretty much like we humans do, at least in the western hemisphere). So we start with `doIt`. Next thing we see is a `()`. That's top priority, so we call the function. After that we encounter the `/`, priority 13. It's followed by another `(`, priority 16. So we evaluate `j=2` first. Having done that, we return to the division. That happens to be the last step in the evaluation. – Stephan Rauh Jan 12 '17 at 22:16
Another possible explanation is that `doIt()` is a singular term. In Java, you can't do anything with the name of a method. Other programming languages may treat it as an useful entity of its own (function pointers, lambda expression etc.). Java doesn't do such a thing. If it finds a method name in an expression, the only sensible thing to do is to execute it. From this point of view, `doIt` and `()` belong together and can't be separated. They are a single, atomar term of the expression. Little wonder the method is evaluated first, no matter what comes next. – Stephan Rauh Jan 12 '17 at 22:21

confused with return value of assignment operation in java

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