mod of a negative number

[gk]

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

how this is working ?

 

[Lars Enderin]

gk skrev:

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

how this is working ?

(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.

 

[Patricia Shanahan]

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

how this is working ?

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

 

[gk]

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.

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.

 

[Richard F.L.R.Snashall]

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.

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

 

[Lee Weiner]

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

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

 

[Patricia Shanahan]

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

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

 

[gk]

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

nice explanation!

 

[kstahmer]

gk wrote:
> Int a = -5; Int b = -2;
> System.out.println(a % b); // -1
>
> how this is working ?
>

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,

Patricia

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.