[SOLVED] BASH

GPGAgent · 12-29-2018, 10:45 AM

I need a division to always return an even number

I've tried (dividend+divisor-1)/divisor and 1+(dividend-1)/divisor with no luck

Code:

onk@XEON4 ~/LK/PROCESS $ dividend=1400
onk@XEON4 ~/LK/PROCESS $ divisor=3
onk@XEON4 ~/LK/PROCESS $ echo $dividend $divisor $[(dividend+divisor-1)/divisor] $[1+(dividend-1)/divisor]
1400 3 467 467

It must be easy enough, isn't it? is there a bash function?

pan64 · 12-29-2018, 10:54 AM

$[ ] is deprecated in bash. Instead you can try $(( arithmetic expression )). See man bash about it.
try:

Code:

v=$(( dividend/2/divisor ))
echo $(( v*2 ))

if I understand it well

Turbocapitalist · 12-29-2018, 10:56 AM

Have you tried modulus?

Code:

r=3; 
echo $(($r-$r%2));

GPGAgent · 12-29-2018, 11:13 AM

Quote:

Originally Posted by Turbocapitalist

Have you tried modulus?

Code:

r=3; 
echo $(($r-$r%2));

Nope:

Code:

onk@XEON4 ~/LK/PROCESS $ r=1400
onk@XEON4 ~/LK/PROCESS $ echo $(($r-$r%3));
1398

GPGAgent · 12-29-2018, 11:15 AM

Quote:

Originally Posted by pan64

$[ ] is deprecated in bash. Instead you can try $(( arithmetic expression )). See man bash about it.
try:

Code:

v=$(( dividend/2/divisor ))
echo $(( v*2 ))

if I understand it well

Looks good to me and so obvious

Code:

onk@XEON4 ~/LK/PROCESS $ echo $dividend $divisor echo $(( dividend/2/divisor*2 )) $(( dividend/divisor ))
1401 3 echo 466 467

michaelk · 12-29-2018, 11:16 AM

Quote:

It must be easy enough, isn't it?

No. Even or odd only applies to integers and in your example the quotient is not an integer i.e 1400/3=466.67... I'm not much of a mathematician so according to wikipedia

"when the quotient is an integer, it will be even if and only if the dividend has more factors of two than the divisor."

https://en.wikipedia.org/wiki/Parity_(mathematics)

NevemTeve · 12-29-2018, 11:18 AM

Maybe this (with integer aritmethic): (p+q)/(2q)*2

Turbocapitalist · 12-29-2018, 11:20 AM

modulus would go like this:

Code:

dividend=1400; 
divisor=7; 
r=$((dividend/divisor)); 
echo  $dividend $divisor $(($r-$r%2));

The other solutions might be better.

pan64 · 12-29-2018, 11:22 AM

Quote:

Originally Posted by GPGAgent

Nope:

Code:

onk@XEON4 ~/LK/PROCESS $ r=1400
onk@XEON4 ~/LK/PROCESS $ echo $(($r-$r%3));
1398

I think you missed:
$r is the result $(( dividend/divisor )) and you can use $(( r-r%2 )) to make it even. This 2 is 2 in any case, not related to divisor or anything else.

ehartman · 12-29-2018, 11:29 AM

Quote:

Originally Posted by michaelk

No. Even or odd only applies to integers and in your example the quotient is not an integer i.e 1400/3=466.67...

You forget one thing: arithmetic expressions in bash are always calculated as integers

Code:

Evaluation is done in fixed-width integers with no check for overflow, though
division by 0 is trapped and flagged as an error.

(from the bash man page under ARITHMETIC EVALUATION).

So in bash $(( 1400 / 3 )) will return 466 without any fractional part.
I tested it to be sure and indeed, 466 was the answer.

GPGAgent · 12-29-2018, 11:40 AM

Quote:

Originally Posted by pan64

I think you missed:
$r is the result $(( dividend/divisor )) and you can use $(( r-r%2 )) to make it even. This 2 is 2 in any case, not related to divisor or anything else.

Sorry, misread your reply, and yes that works just fine:

Code:

onk@XEON4 ~/LK/PROCESS $ echo $dividend $divisor $((dividend/divisor - dividend/divisor%2)) 
1401 3 466

10 out of 10

GPGAgent · 12-29-2018, 11:42 AM

Thanks to the others as well, modulus is good but I'm going with divide by 2 then multiply by 2 - the most obvious and good enough for my usage.

Happy New year everyone

GPGAgent · 12-29-2018, 12:00 PM

Quote:

Originally Posted by michaelk

No. Even or odd only applies to integers and in your example the quotient is not an integer i.e 1400/3=466.67... I'm not much of a mathematician....

True enough, but bash only works in integer values ;-)

Code:

onk@XEON4 ~/LK/PROCESS $ echo $dividend $divisor $((dividend/divisor)) 
1400 3 466

Thanks tho', and happy new year

l0f4r0 · 12-29-2018, 06:25 PM

Obviously you got what you were searching for because your thread is marked as [SOLVED] now but:

do you want to allow division only if the result is even?
OR
do you want to process all divisions and get an even result?

I'm asking this question because I can't really see a useful application for case #2 (considering the fact that your result can sometimes be mathematically wrong)...
Out of curiosity, could you share how all of this could be useful to you?

ehartman · 12-29-2018, 08:50 PM

Quote:

Originally Posted by l0f4r0

I'm asking this question because I can't really see a useful application for case #2 (considering the fact that your result can sometimes be mathematically wrong)...

As bash arithmetic is integer only, just the result from 1400 / 3 already is mathematically wrong, although you can get the remainder with the % operator.
The result of above division will be 466, with remainder 2 (which both happen to be even).