Calculate Java Int Overflow

Question

Is there a formula to calculate what the overflow of a Java int would be?

Example: if I add 1 to Integer.MAX_VALUE; the answer is not 2147483648, but rather -2147483648.

Question: if I wanted to calculate what Java would print for a value larger than 2^32, is there an easy mathematical expression (theoretical, not in code)?

@BoristheSpider how would you translate (INTEGER.MAX_VALUE * 10 - INTEGER.MIN_VALUE) * randomNumber;? — Luiggi Mendoza
– Luiggi Mendoza, Commented May 5, 2014 at 20:23
@Luiggi honestly; I would plug it into a main method and print the output. — Boris the Spider
– Boris the Spider, Commented May 5, 2014 at 20:24
But is there an expression (not in code) that is just algebraic based. Example: +/-(2147483648 - |int|)? — Jake Chasan
– Jake Chasan, Commented May 5, 2014 at 20:26

ajb · Accepted Answer · 2014-05-05 20:52:28Z

5

((x + 2³¹) mod 2³²) - 2³¹

Is this what you're looking for? That should be the result of any mathematical operation on a machine that uses 32-bit signed 2's complement integers. That is, if the mathematical value of an operation returns x, the above formula gives the integer that would actually be stored (if the operation doesn't fault, and it's not a "saturating" operation).

Note that I'm using "mod" with a mathematical definition, not the way the % operator works in Java or C. That is, A mod B, where A and B are integers and B > 0, always returns an integer in the range 0 .. B-1, e.g. (-1) mod 5 = 4. More specifically, A mod B = A - B*floor(A/B).

edited May 5, 2014 at 20:52

answered May 5, 2014 at 20:45

ajb

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2 Comments

Display name Over a year ago

Let me see if I have this straight. For nonnegative values of x <= 2^31-1, the addition operation in the formula corresponds to flipping the 31st bit from 0 to 1, the result of the modulo operator is the same value, and then subtracting 2^31 flips the 31st bit back to 0. For the other cases where x>2^31-1 or when x is negative, what is going on with the binary representation of x during this operation?

ajb Over a year ago

@NicholasCousar This is intended to be a mathematical formula for what the result of the computer's operation will be. It is not intended to describe what the computer actually does with the binary representation of the number. Trying to think of this formula in terms of a binary representation will probably just create confusion.

Buzz · Accepted Answer · 2014-05-05 20:56:16Z

4

In java an int is 32 bits but it is also signed, which means that the first bit is the "negative" sign. 1 means negative and 0 means positive. Because of this, the largest number is 2147483647 (0111 1111 1111 1111 1111 1111 1111 1111). If you add 1 it makes it 1000 0000 0000 0000 0000 0000 0000 0000 which translates to -2147483648. For any values larger than that you would need to use a long

answered May 5, 2014 at 20:56

Buzz

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Comments

Kyranstar · Accepted Answer · 2014-05-05 20:45:21Z

0

I believe this would work:

 int expected = ((val + Integer.MAX_VALUE) % Integer.MAX_VALUE) - Integer.MAX_VALUE;

answered May 5, 2014 at 20:45

Kyranstar

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1 Comment

ajb Over a year ago

On the right track, but not correct. MAX_VALUE is 2^31 - 1 and you definitely don't want to use that with your modulo operator.

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