Postincrement and preincrement in Java expressions

Question

What does each expression evaluate to? Assume x is 2 before each one.

int num = x++ * 3;
So this would be equivalent to (2)*3 or num=6 and x is now 3.
num *= x;
num =2*2 or 4
(x < 2) && (x > 1)
Becomes FALSE, because (2<2)=false and (2>1)=true so it's false.
(++x < 2) || (x < 1)
(3<2) is false and then ((2+1)<1) is also false, so it's false?
One question is in this case, is the preincrement applied to the variable before the break? Should the second x value be 3 or 2?
I also have the same question for postincrement. Let's say I have num=x++ *x++ where initial x=2. So is this 2*2 or 2*3?

On the fourth one. Since the preincrement is in the first bracket thing (im sure it has a special name) does it effect the value of the x in the second parenthesis comparison? — Chris, Mar 01 '12 at 01:45
Like if the value of x is pre/post incremented in a first half of an AND or OR statement, does it affect the value of the second half's value? — Chris, Mar 01 '12 at 01:45
Why don't you try it out yourself? --> [Ideone](http://ideone.com) — JRL, Mar 01 '12 at 01:47
This question might help: [Pre & post increment operator behavior in C, C++, Java, & C#](http://stackoverflow.com/questions/6457130/pre-post-increment-operator-behavior-in-c-c-java-c-sharp). — Felix Kling, Mar 01 '12 at 01:48
I don't have admin rights to install Eclipse nor the SDK, but thanks to JRL I can do my Java testing now. Really cool program! — Chris, Mar 01 '12 at 01:57

score 1 · Accepted Answer · answered Mar 01 '12 at 01:46

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It's incremented before the "break" yes. Basically it's the first thing java does (parenthesis are still the first ones actually). So in (++x < 2) || (++x < 3) the 2nd ++x happens after the first one if it isn't true.

answered Mar 01 '12 at 01:46

fbernardo

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I realize it adds one to the value of the x for the first half before executing the statement. Let me try to explain this a different way: `var x=2;`
`(x++ < 4)&&(x==2);`
Is the second statement true or false. I am wondering whether preincrement affects the entire line's x value.. – Chris Mar 01 '12 at 01:48
@Chris: Why don't you just run it? If the result is `false`, the second statement is `false`. If it is `true`, it's `true`. That's easy to find out. – Felix Kling Mar 01 '12 at 01:50
@Felix Kling: I don't have admin rights to the comptuer I am on to install either Eclipse or the SDK. Although someone posted this cool online java runner thing I am using now. I've never done it before, so how can you return the value of the OR statement without writing a large if else thing? – Chris Mar 01 '12 at 01:56
@fbernardo if the program gives me this error: "cannot return a value from method whose result type is void" that means the statement is false right? Or does it mean something is actually wrong? – Chris Mar 01 '12 at 02:12
If you declare a void method (like main) you can't return a value. That's what a void method is, a method that doesn't return anything... So, that doesn't mean it's true or false. – fbernardo Mar 01 '12 at 02:13
Btw, if you want to return the result of a boolean expression like "a | b" the return type is "boolean". – fbernardo Mar 01 '12 at 02:15

Jerome · Answer 2 · 2012-03-01T02:34:52.903

int num = x++ * 3; => OK: x = 3, num = 6
num *= x; => what's the initial value of num ? if num = 2, then you are also OK.
(x < 2) && (x > 1) is false when x = 2, OK
(++x < 2) || (x < 1) is false when x = 2, OK as well

I remember I had a look at the openJDK, especially the Lower class in javac source and ++x is translated into x += 1 therefore you can see it as:
((x += 1)) < 2) which is (whithout type casting): ((x = (x + 1)) < 2). The second test will have the newer x value i.e 3 because java evaluates the conditions from left to right.

score 0 · Answer 3 · answered Mar 01 '12 at 01:49

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Yes those are some very basic concepts, do yourself a favour. Download unittest framework something like jUnit would do. Then you can run all these tests quickly with a few asserts that will show you exactly what is going on. Another way to do it without much overhead is just to print it out on the console.

int num = x++ * 3;
System.out.println( "num=" + num );  // num=6

answered Mar 01 '12 at 01:49

Orn Kristjansson

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I don't have admin rights to the comptuer I am on to install either Eclipse or the SDK. Although someone posted this cool online java runner thing I am using now. – Chris Mar 01 '12 at 02:00

score 0 · Answer 4 · answered Mar 01 '12 at 01:51

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num *= x;
num =2*2 or 4

I don't know where you got that.

num *= x; is equivalent to num = num * x;. If num was 6 (from the previous statement) then it would now be 12 (assuming x is still 2).

answered Mar 01 '12 at 01:51

Hot Licks

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in this case, x becomes 3 after the first sum, so num will be 18. – Jeremy D Mar 01 '12 at 01:55
Thats what She gave me. I think shes assuming that all of the programs are separate with the x value initialized at 2. – Chris Mar 01 '12 at 01:59

score 0 · Answer 5 · answered Mar 01 '12 at 01:55

"Assume x is 2 before each one." Is that a given assumption or is it an assumption you're making? I don't believe that you can make that assumption. I believe that you have evaluate all of these statements in sequence.
What is the expanded form of a *= b?
What is the initial value of num when you're at the second statement?
On your last statement you did correctly (assuming your initial conditions are correct) compute (2+1) < 1 because x has been incremented.

Keep track of the values of x and num after each statement and I think you'll get to the right answer.

Good luck!

Postincrement and preincrement in Java expressions

5 Answers5