Does Incrementing In Java Change its Immediate Value?

Asked Dec 09 '18 at 15:20

Active Dec 09 '18 at 15:25

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I have been given some practice questions to do for an upcoming test and there is one that has confused me:

"What is print out by the following segment of code:

int var1 = 2;

int var2 = 6;

int var3 = var1;

var3 = var2++ * var3 + var2 / var3;

System.out.println(var3);"

It looks simple enough, but I cannot understand how the answer is 15, and not 17.5 as I put (and that answer is a float so it isn't valid anyway.)

The only way I see of getting 15 is if, when you do var2++, it does not actually increment var2 to 7. Also, I am not sure if var2++ updates the var2 variable or just increments temporarily for that one instance.

The way I have been doing it is by following the order of presedence...

var3 = ((var2++) * var3) + (var2/var3)

But this gives me...

var3 = (7 * 2) + (7/2)

var3 = 14 + 3.5

var 3 = 17.5

I have run the code in a Java compiler and I am in fact getting 15.

Anyone know why?

edited Dec 09 '18 at 15:25

Gayan Mettananda

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asked Dec 09 '18 at 15:20

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pre vs post increment. `var3 = ++var2 * var3 + var2 / var3;` – Elliott Frisch Dec 09 '18 at 15:25
You're dividing integers, and that is an integer division, which gives an int: 5 / 2 = 2. If you use a double, on the contrary, 5.0 / 2 = 2.5 – JB Nizet Dec 09 '18 at 15:27
@JBNizet `var3` is also an `int`, so storing a `double` is impossible anyway. – Elliott Frisch Dec 09 '18 at 15:27
@ElliottFrisch yes, that's my point. – JB Nizet Dec 09 '18 at 15:28
@JBNizet Just elaborating... I started to say something about the integer math in my original comment, but I got lost in the increment. :) – Elliott Frisch Dec 09 '18 at 15:37
`var3 = 6 * 2 + 7/2 = 12 + 3 = 15` – UninformedUser Dec 09 '18 at 15:38

Does Incrementing In Java Change its Immediate Value?

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