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)?

Answer 1

((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).

Answer 2

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

Answer 3

I believe this would work:

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