Floating point comparison with variable in bash [duplicate] - bash

Question

This question already has answers here:
Bash: Integer expression expected [duplicate]
(2 answers)
Closed 7 years ago.
I want to compare a floating point variable to an integer.
I know this is not the best to do with bash, but my whole script is already written in bash.
$number can be any integer. If it below or equal 50, I want output1, for all others I want an output with the other variable k. This is what I have so far:
number=43
test=$(echo "scale=2; $number/50" | bc -l)
echo "$test"
for k in {1..5}
do
if ["$test" -le 1]
then echo "output"
elif ["$test" -gt $k]
then echo "output$k"
fi
done
If I try with test=0.43, the first loop does not even work. I think it has to do with an integer and a floating point comparison but cannot make it work.
Anything I am missing?
PS:this [0.43: command not found is what the terminal outputs.

Answer 1

Bash can't handle floats. Pipe to bc instead:
if [ $(echo " $test > $k" | bc) -eq 1 ]
The error you see though is because the test command (i.e. the [) needs spaces before and after
It is even better to use (( ... )) since you compare numbers like this:
if (( $(bc <<< "$test > $k") ))
The part in the loop should look like this:
if (( $(bc <<< "$test <= 1") ))
then
echo "output"
elif (( $(bc <<< "$test > $k") ))
then
echo "output$k"
fi
Relational expressions evaluate to 0, if the relation is false, and 1 if the relation is true [source]. Note however that is a behavior of GNU bc, and it is not POSIX compiant.

Answer 2

Kind of an old question, but it bears an additional answer I think.
While piping to a higher precision calculator (bc or dc) works, it is at the cost of a fork and a an extra process, since those calculators are not built in to bash. One thing that IS built in, though, is printf. So if you can deal with your numbers being within a particular number of decimal places, you can "fake" floating point comparisons, with a function like this:
#!/usr/bin/env bash
function [[[ () {
local LANG=C lhs rhs
printf -v lhs '%07.3f' "$1"; lhs=${lhs/./}
printf -v rhs '%07.3f' "$3"; rhs=${rhs/./}
case "$2" in
-lt) return $(( ! ( 10#$lhs < 10#$rhs ) )) ;;
-le) return $(( ! ( 10#$lhs <= 10#$rhs ) )) ;;
-eq) return $(( ! ( 10#$lhs == 10#$rhs ) )) ;;
-ge) return $(( ! ( 10#$lhs >= 10#$rhs ) )) ;;
-gt) return $(( ! ( 10#$lhs > 10#$rhs ) )) ;;
esac
}
number=${1:-43}
test=$(dc -e "2k $number 50 / p")
echo "$test"
for k in {1..5}; do
if [[[ "$test" -le 1 ]]]; then
echo "output"
elif [[[ "$test" -gt "$k" ]]]; then
echo "output $k"
fi
done
A few things to consider here.
I've named the function [[[ to be cute. You can name it whatever you like. ntest or mynumericcomparison or even [[[.
printf is an internal function within bash, so despite the fact that it's on your path, it doesn't cost a fork.
As it stands, the function supports numbers up to 999.999. If you need higher numbers (or more precision), adjust the printf formats.
The 10# at the beginning of each variable inside the case statement is to force the comparison to happen at base 10, since a zero-padded number might otherwise be interpreted as octal.
See also: http://mywiki.wooledge.org/BashFAQ/022