Strange behaviour on dividing -ve int/-ve int ,in java

Question

Although I am not new to java, I observed this peculiar behavior the other day. I was refreshing my basics by running code consisting of basic arithmetic operations. Now according to java (and basic rules of arithmetic's ofcourse) , -ve * -ve OR -ve / -ve is a +ve number.

But compiling this source:-

int b = Integer.MIN_VALUE / -1;
System.out.println("b:  " + b);

Gives me output:-

b: -2147483648

Which is -ve, Can anyone point me whats wrong? I know it must be small thing that I cant notice.

```-2147483648 / -1``` => ```2147483648``` => overflow => ```-2147483648``` — Mateusz Dymczyk, Sep 17 '14 at 07:55
Related: http://stackoverflow.com/questions/9397475/why-integer-max-value-1-integer-min-value — Maroun, Sep 17 '14 at 07:56
funny fact: a good IDE will tell you about the overflow (just checked) — Mateusz Dymczyk, Sep 17 '14 at 07:57
@MateuszDymczyk thanks , that was simple :D , but my IDE gave no warning. (eclipse kepler sr2) — Mustafa sabir, Sep 17 '14 at 08:36

score 10 · Accepted Answer · answered Sep 17 '14 at 07:55

To divide by -1 is the same as negating the number.

Since the range of integers (-2147483648 to 2147483647) is 1 larger in magnitude on the negative side -Integer.MIN_VALUE equals Integer.MAX_VALUE+1 which overflows back to Integer.MIN_VALUE.

System.out.println(Integer.MIN_VALUE == -Integer.MIN_VALUE); // prints 'true'

score 2 · Answer 2 · answered Sep 17 '14 at 08:00

You have an overflow. Per java specification, the sign is not guaranteed in overflow.

The overflow is due to the fact that

Integer.MIN_VALUE = -2147483648;
Integer.MAX_VALUE = 2147483647;

so -2147483648*(-1) is 2147483648, which is over the max value.

Strange behaviour on dividing -ve int/-ve int ,in java

2 Answers2