short-circuit question

I think,

1) a = true is evaluated, so a = true
2) b = true is not evaluated, so b = false (short-circuit)
3) the original expresion would be, boolean x = true && (c = true)
4) c = true is evaluated, so c = true
5) boolean x = true

But the answer is, a = true, b = false, c = false
Any ideas?
Thanks a lot.

Hey Grady,

It actually works this way.

op1 || op2 where op1, op2 are two operands.

if op1 is "true" then op2 will not be evaluated, EVEN IF op2 is an expression (in this case (b=true) && (c=true)).
Similarly for &&, if op1 is false then op2 is not evaluated at all.

One more point to note is both ||, && have same level of precedence and so if they appear in a expression, they will be evaluated as usually (from left-right).

Say,

Here it will be evaluated as (false || (true && (true || false))) and so will print "check precedence".

Hope that was helpfull,
Kits

Hi Grady,

If you look at operator precedence order of || and &&, && has the higher precedence than operator ||.
And also if you remember java always evaluates left oprand first for binary operators.
So the expression will get evaluated as -

boolean x = (a = true) || (b = true) && (c = true);//Original expression.

x = ((a = true) || ((b = true) && (c = true)));//According to the precedence.

x = (true || ((b = true) && (c = true)));//Left oprand evaluated first.

At this point sencond expression i.e ((b = true) && (c = true)) will never get evaluated as has got short circuited.

Hence the result is,

x = true; with a = true , b = false & c = false.

I hope this answers your question and not adding any confusion

Correct me if my understanding is wrong!

Regards,
Kapil

Thank you, guys.
I'm quite clear how it works after you two explained.