# mod of a negative number

Discussion in 'Java' started by gk, Nov 5, 2006.

1. ### gkGuest

Int a = -5; Int b = -2;
System.out.println(a % b); // -1

how this is working ?

gk, Nov 5, 2006

2. ### Lars EnderinGuest

gk skrev:
(How does this work?)
According to the definition of the % operator -1 is the rest after
taking out a multiple (-4) of -2 from -5.

Lars Enderin, Nov 5, 2006

3. ### Patricia ShanahanGuest

Strictly speaking, there is no "mod" operator in Java. % is defined to
be "remainder". It is the same as modulo for positive operands.

For integer operands, % is designed to maintain the identity:
(a/b)*b+(a%b) is equal to a.

Java integer division rounds towards zero, so -5/-2 is 2.

( (-5)/(-2) ) * (-2) + (-1) is -5.

See the JLS,
http://java.sun.com/docs/books/jls/second_edition/html/expressions.doc.html#239829

Patricia

Patricia Shanahan, Nov 5, 2006
4. ### gkGuest

true.....but -2<-1 , so we can divide it further theoretically ...is
not it ? hmm...but if we keep on dividing its going to be an infinite
loop.

gk, Nov 5, 2006
5. ### Richard F.L.R.SnashallGuest

A nitpick question: Is the division ( (8)/(3) ) also 2 in Java?
If so, why is this be called "rounding"?

Richard F.L.R.Snashall, Nov 5, 2006
6. ### Lee WeinerGuest

It is not called rounding. It is called integer division.

Lee Weiner, Nov 5, 2006
7. ### Patricia ShanahanGuest

I usually use "rounding" in the sense in which it is used in e.g. the
IEEE 754 standard, to mean modifying the infinitely precise result of a
calculation to fit in the destination's format.

In that usage, it includes rounding directed rounding, such as rounding
towards zero, as well as the various flavors of round to nearest. Some
people use "truncation" when the rounding is towards either zero or
negative infinity.

Patricia

Patricia Shanahan, Nov 5, 2006
8. ### gkGuest

nice explanation!

gk, Nov 5, 2006
9. ### kstahmer

Joined:
May 13, 2010
Messages:
1
0
Illuminating post - thanks.

The JLS example was also illuminating:

5 % 3 produces 2
5 % (-3) produces 2
(-5) % 3 produces -2
(-5) % (-3) produces -2

This got me thinking...

Suppose a and b are integers.

We have two special cases:
1. If b == 0, then a % b is NaN (JLS specification).
2. If nonzero b divides a, then a % b == 0 (In particular, every nonzero b divides 0, so 0 % b == 0).
Which leaves us with nonzero b does not divide a.

Then

a % b > 0 if a > 0

and

a % b < 0 if a < 0.

Hence, if nonzero b does not divide a, then the sign of a % b equals the sign of a.

You can see this formally by noting:

(a / b) * b + a % b == a

implies

a % b == a - (a / b) * b

and if nonzero b does not divide a, then

Math.abs((a / b) * b) < Math.abs(a),

due integer division rounding towards zero.

kstahmer, May 13, 2010