Get ceiling integer from number in linux (BASH)

Question

How would I do something like:

ceiling(N/500)

N representing a number.

But in a linux Bash script

Answer 1

Some more concise Awk logic

awk '
function ceil(ip) {
  print ip%1 ? int(ip)+1 : ip
}  
BEGIN {
  ceil(1000/500)
  ceil(1001/500)
}
'

Result

2
3

Answer 2

Why use external script languages? You get floor by default. To get ceil, do

$ divide=8; by=3; (( result=(divide+by-1)/by )); echo $result
3
$ divide=9; by=3; (( result=(divide+by-1)/by )); echo $result
3
$ divide=10; by=3; (( result=(divide+by-1)/by )); echo $result
4
$ divide=11; by=3; (( result=(divide+by-1)/by )); echo $result
4
$ divide=12; by=3; (( result=(divide+by-1)/by )); echo $result
4
$ divide=13; by=3; (( result=(divide+by-1)/by )); echo $result
5
....

To take negative numbers into account you can beef it up a bit. Probably cleaner ways out there but for starters

$ divide=-10; by=10; neg=; if [ $divide -lt 0 ]; then (( divide=-divide )); neg=1; fi; (( result=(divide+by-1)/by )); if [ $neg ]; then (( result=-result )); fi; echo $result
-1

$ divide=10; by=10; neg=; if [ $divide -lt 0 ]; then (( divide=-divide )); neg=1; fi; (( result=(divide+by-1)/by )); if [ $neg ]; then (( result=-result )); fi; echo $result
1

(Edited to switch let ... to (( ... )).)

Answer 3

If you are already familiar with the Python library, then rather than learn bc, you might want to define this bash function:

pc () { pyexpr="from math import *; print($@)"; python -c "$pyexpr"; }

Then:

pc "ceil(3/4)"
1

but also any valid python expression works:

pc pi / 4
0.7853981633974483

pc "'\n'.join(['Pythagoras said that %3.2f^2 + %3.2f^2 is always %3.2f'
    % (sin(ai), cos(ai), sin(ai)**2 + cos(ai)**2)
    for ai in [pi / 4 * k for k in range(8)]])"
Pythagoras said that 0.00^2 + 1.00^2 is always 1.00
Pythagoras said that 0.71^2 + 0.71^2 is always 1.00
Pythagoras said that 1.00^2 + 0.00^2 is always 1.00
Pythagoras said that 0.71^2 + -0.71^2 is always 1.00
Pythagoras said that 0.00^2 + -1.00^2 is always 1.00
Pythagoras said that -0.71^2 + -0.71^2 is always 1.00
Pythagoras said that -1.00^2 + -0.00^2 is always 1.00
Pythagoras said that -0.71^2 + 0.71^2 is always 1.00

Answer 4

Using the gorgeous 'printf' 1 will round up to the next integer

printf %.0f $float
or
printf %.0f `your calculation formula`
or
printf %.0f $(your calculation formula)

ref: how to remove decimal from a variable?

Answer 5

Floor () {
  DIVIDEND=${1}
  DIVISOR=${2}
  RESULT=$(( ( ${DIVIDEND} - ( ${DIVIDEND} % ${DIVISOR}) )/${DIVISOR} ))
  echo ${RESULT}
}
R=$( Floor 8 3 )
echo ${R}

Ceiling () {
  DIVIDEND=${1}
  DIVISOR=${2}
  $(( ( ( ${DIVIDEND} - ( ${DIVIDEND} % ${DIVISOR}) )/${DIVISOR} ) + 1 ))
  echo ${RESULT}
}
R=$( Ceiling 8 3 )
echo ${R}

Answer 6

Without specifying any function, we can use the following awk script:

echo x y | awk '{ r=$1 % $2; q=$1/y; if (r != 0) q=int(q+1); print q}'

Not sure this one get any logical error or not. Please correct.