Multiplying integer and float variables in bash [duplicate] - bash

Question

I am trying to divide two image widths in a Bash script, but bash gives me 0 as the result:
RESULT=$(($IMG_WIDTH/$IMG2_WIDTH))
I did study the Bash guide and I know I should use bc, in all examples in internet they use bc. In echo I tried to put the same thing in my SCALE but it didn't work.
Here is the example I found in the tutorials:
echo "scale=2; ${userinput}" | bc
How can I get Bash to give me a float like 0.5?

Answer 1

You can't. bash only does integers; you must delegate to a tool such as bc.

Answer 2

you can do this:
bc <<< 'scale=2; 100/3'
33.33
UPDATE 20130926 : you can use:
bc -l <<< '100/3' # saves a few hits
33.33333333333333333333

Answer 3

bash
As noted by others, bash does not support floating point arithmetic, although you could fake it with some fixed decimal trickery, e.g. with two decimals:
echo $(( 100 * 1 / 3 )) | sed -e 's/..$/.&/;t' -e 's/.$/.0&/'
Output:
.33
See Nilfred's answer for a similar but more concise approach.
Alternatives
Besides the mentioned bc and awk alternatives there are also the following:
clisp
clisp -x '(/ 1.0 3)'
with cleaned up output:
clisp --quiet -x '(/ 1.0 3)'
or through stdin:
echo '(/ 1.0 3)' | clisp --quiet | tail -n1
dc
echo 2k 1 3 /p | dc
genius cli calculator
echo 1/3.0 | genius
ghostscript
echo 1 3 div = | gs -dNODISPLAY -dQUIET | sed -n '1s/.*>//p'
gnuplot
echo 'pr 1/3.' | gnuplot
Imagemagick
convert xc: -format '%[fx:1/3]' info:
or through stdin:
echo 1/3 | { convert xc: -format "%[fx:$(cat)]" info:; }
jq
jq -n 1/3
or through stdin:
echo 1/3 | jq -nf /dev/stdin
ksh
echo 'print $(( 1/3. ))' | ksh
lua
lua -e 'print(1/3)'
or through stdin:
echo 'print(1/3)' | lua
maxima
echo '1/3,numer;' | maxima
with cleaned up output:
echo '1/3,numer;' | maxima --quiet | sed -En '2s/[^ ]+ [^ ]+ +//p'
node
echo 1/3 | node -p
octave
echo 1/3 | octave
perl
echo print 1/3 | perl
python2
echo print 1/3. | python2
python3
echo 'print(1/3)' | python3
R
echo 1/3 | R --no-save
with cleaned up output:
echo 1/3 | R --vanilla --quiet | sed -n '2s/.* //p'
ruby
echo puts 1/3.0 | ruby
units
units 1/3
with compact output:
units --com 1/3
wcalc
echo 1/3 | wcalc
with cleaned up output:
echo 1/3 | wcalc | tr -d ' ' | cut -d= -f2
zsh
print $(( 1/3. ))
or through stdin:
echo 'print $(( 1/3. ))' | zsh
Other sources
Stéphane Chazelas answered a similar question over on UL.

Answer 4

Improving a little the answer of marvin:
RESULT=$(awk "BEGIN {printf \"%.2f\",${IMG_WIDTH}/${IMG2_WIDTH}}")
bc doesn't come always as installed package.

Answer 5

You could use bc by the -l option (the L letter)
RESULT=$(echo "$IMG_WIDTH/$IMG2_WIDTH" | bc -l)

Answer 6

As an alternative to bc, you can use awk within your script.
For example:
echo "$IMG_WIDTH $IMG2_WIDTH" | awk '{printf "%.2f \n", $1/$2}'
In the above, " %.2f " tells the printf function to return a floating point number with two digits after the decimal place. I used echo to pipe in the variables as fields since awk operates properly on them. " $1 " and " $2 " refer to the first and second fields input into awk.
And you can store the result as some other variable using:
RESULT = `echo ...`

Answer 7

Well, before float was a time where fixed decimals logic was used:
IMG_WIDTH=100
IMG2_WIDTH=3
RESULT=$((${IMG_WIDTH}00/$IMG2_WIDTH))
echo "${RESULT:0:-2}.${RESULT: -2}"
33.33
Last line is a bashim, if not using bash, try this code instead:
IMG_WIDTH=100
IMG2_WIDTH=3
INTEGER=$(($IMG_WIDTH/$IMG2_WIDTH))
DECIMAL=$(tail -c 3 <<< $((${IMG_WIDTH}00/$IMG2_WIDTH)))
RESULT=$INTEGER.$DECIMAL
echo $RESULT
33.33
The rationale behind the code is: multiply by 100 before divide to get 2 decimals.

Answer 8

It's perfect time to try zsh, an (almost) bash superset, with many additional nice features including floating point math. Here is what your example would be like in zsh:
% IMG_WIDTH=1080
% IMG2_WIDTH=640
% result=$((IMG_WIDTH*1.0/IMG2_WIDTH))
% echo $result
1.6875
This post may help you: bash - Worth switching to zsh for casual use?

Answer 9

If you found the variant of your preference you can also wrap it into a function.
Here I'm wrapping some bashism into a div function:
One liner:
function div { local _d=${3:-2}; local _n=0000000000; _n=${_n:0:$_d}; local _r=$(($1$_n/$2)); _r=${_r:0:-$_d}.${_r: -$_d}; echo $_r;}
Or multi line:
function div {
local _d=${3:-2}
local _n=0000000000
_n=${_n:0:$_d}
local _r=$(($1$_n/$2))
_r=${_r:0:-$_d}.${_r: -$_d}
echo $_r
}
Now you have the function
div <dividend> <divisor> [<precision=2>]
and use it like
> div 1 2
.50
> div 273 123 5
2.21951
> x=$(div 22 7)
> echo $x
3.14
UPDATE
I added a little script which provides you the basic operations with floating point numbers for bash:
Usage:
> add 1.2 3.45
4.65
> sub 1000 .007
999.993
> mul 1.1 7.07
7.7770
> div 10 3
3.
> div 10 3.000
3.333
And here the script:
#!/bin/bash
__op() {
local z=00000000000000000000000000000000
local a1=${1%.*}
local x1=${1//./}
local n1=$((${#x1}-${#a1}))
local a2=${2%.*}
local x2=${2//./}
local n2=$((${#x2}-${#a2}))
local n=$n1
if (($n1 < $n2)); then
local n=$n2
x1=$x1${z:0:$(($n2-$n1))}
fi
if (($n1 > $n2)); then
x2=$x2${z:0:$(($n1-$n2))}
fi
if [ "$3" == "/" ]; then
x1=$x1${z:0:$n}
fi
local r=$(($x1"$3"$x2))
local l=$((${#r}-$n))
if [ "$3" == "*" ]; then
l=$(($l-$n))
fi
echo ${r:0:$l}.${r:$l}
}
add() { __op $1 $2 + ;}
sub() { __op $1 $2 - ;}
mul() { __op $1 $2 "*" ;}
div() { __op $1 $2 / ;}

Answer 10

How to do floating point calculations in bash:
Instead of using "here strings" (<<<) with the bc command, like one of the most-upvoted examples does, here's my favorite bc floating point example, right from the EXAMPLES section of the bc man pages (see man bc for the manual pages).
Before we begin, know that an equation for pi is: pi = 4*atan(1). a() below is the bc math function for atan().
This is how to store the result of a floating point calculation into a bash variable--in this case into a variable called pi. Note that scale=10 sets the number of decimal digits of precision to 10 in this case. Any decimal digits after this place are truncated.
pi=$(echo "scale=10; 4*a(1)" | bc -l)
Now, to have a single line of code that also prints out the value of this variable, simply add the echo command to the end as a follow-up command, as follows. Note the truncation at 10 decimal places, as commanded:
pi=$(echo "scale=10; 4*a(1)" | bc -l); echo $pi
3.1415926532
Finally, let's throw in some rounding. Here we will use the printf function to round to 4 decimal places. Note that the 3.14159... rounds now to 3.1416. Since we are rounding, we no longer need to use scale=10 to truncate to 10 decimal places, so we'll just remove that part. Here's the end solution:
pi=$(printf %.4f $(echo "4*a(1)" | bc -l)); echo $pi
3.1416
Here's another really great application and demo of the above techniques: measuring and printing run-time.
(See also my other answer here).
Note that dt_min gets rounded from 0.01666666666... to 0.017:
start=$SECONDS; sleep 1; end=$SECONDS; dt_sec=$(( end - start )); dt_min=$(printf %.3f $(echo "$dt_sec/60" | bc -l)); echo "dt_sec = $dt_sec; dt_min = $dt_min"
dt_sec = 1; dt_min = 0.017
Related:
[my answer] https://unix.stackexchange.com/questions/52313/how-to-get-execution-time-of-a-script-effectively/547849#547849
[my question] What do three left angle brackets (`<<<`) mean in bash?
https://unix.stackexchange.com/questions/80362/what-does-mean/80368#80368
https://askubuntu.com/questions/179898/how-to-round-decimals-using-bc-in-bash/574474#574474

Answer 11

While you can't use floating point division in Bash you can use fixed point division. All that you need to do is multiply your integers by a power of 10 and then divide off the integer part and use a modulo operation to get the fractional part. Rounding as needed.
#!/bin/bash
n=$1
d=$2
# because of rounding this should be 10^{i+1}
# where i is the number of decimal digits wanted
i=4
P=$((10**(i+1)))
Pn=$(($P / 10))
# here we 'fix' the decimal place, divide and round tward zero
t=$(($n * $P / $d + ($n < 0 ? -5 : 5)))
# then we print the number by dividing off the interger part and
# using the modulo operator (after removing the rounding digit) to get the factional part.
printf "%d.%0${i}d\n" $(($t / $P)) $(((t < 0 ? -t : t) / 10 % $Pn))

Answer 12

I know it's old, but too tempting. So, the answer is: you can't... but you kind of can. let's try this:
$IMG_WIDTH=1024
$IMG2_WIDTH=2048
$RATIO="$(( IMG_WIDTH / $IMG2_WIDTH )).$(( (IMG_WIDTH * 100 / IMG2_WIDTH) % 100 ))"
Like that, you get 2 digits after the point, truncated (call it rounding to the lower) in pure bash (no need to launch other processes). Of course, if you only need one digit after the point you multiply by 10 and do modulo 10.
What this does:
first $((...)) does integer division;
second $((...)) does integer division on something 100 times larger, essentially moving your 2 digits to the left of the point, then (%) getting you only those 2 digits by doing modulo.
Bonus track: bc version × 1000 took 1.8 seconds on my laptop, while the pure bash one took 0.016 seconds.

Answer 13

It's not really floating point, but if you want something that sets more than one result in one invocation of bc...
source /dev/stdin <<<$(bc <<< '
d='$1'*3.1415926535897932384626433832795*2
print "d=",d,"\n"
a='$1'*'$1'*3.1415926535897932384626433832795
print "a=",a,"\n"
')
echo bc radius:$1 area:$a diameter:$d
computes the area and diameter of a circle whose radius is given in $1

Answer 14

There are scenarios in wich you cannot use bc becouse it might simply not be present, like in some cut down versions of busybox or embedded systems. In any case limiting outer dependencies is always a good thing to do so you can always add zeroes to the number being divided by (numerator), that is the same as multiplying by a power of 10 (you should choose a power of 10 according to the precision you need), that will make the division output an integer number. Once you have that integer treat it as a string and position the decimal point (moving it from right to left) a number of times equal to the power of ten you multiplied the numerator by. This is a simple way of obtaining float results by using only integer numbers.

Answer 15

For those trying to calculate percentages with the accepted answer, but are losing precision:
If you run this:
echo "scale=2; (100/180) * 180" | bc
You get 99.00 only, which is losing precision.
If you run it this way:
echo "result = (100/180) * 180; scale=2; result / 1" | bc -l
Now you get 99.99.
Because you're scaling only at the moment of printing.
Refer to here

Answer 16

Use calc. It's the easiest I found
example:
calc 1+1
2
calc 1/10
0.1

Answer 17

** Injection-safe floating point math in bash/shell **
Note: The focus of this answer is provide ideas for injection-safe solution to performing math in bash (or other shells). Of course, same can be used, with minor adjustment to perform advanced string processing, etc.
Most of the solution that were by presented, construct small scriptlet on the fly, using external data (variables, files, command line, environment variables). The external input can be used to inject malicious code into the engine, many of them
Below is a comparison on using the various language to perform basic math calculation, where the result in floating point. It calculates A + B * 0.1 (as floating point).
All solution attempt avoid creating dynamic scriptlets, which are extremely hard to maintain, Instead they use static program, and pass parameters into designated variable. They will safely handle parameters with special characters - reducing the possibility of code injection. The exception is 'BC' which does not provide input/output facility
The exception is 'bc', which does not provide any input/output, all the data comes via programs in stdin, and all output goes to stdout. All calculation are executing in a sandbox, which does not allow side effect (opening files, etc.). In theory, injection safe by design!
A=5.2
B=4.3
# Awk: Map variable into awk
# Exit 0 (or just exit) for success, non-zero for error.
#
awk -v A="$A" -v B="$B" 'BEGIN { print A + B * 0.1 ; exit 0}'
# Perl
perl -e '($A,$B) = #ARGV ; print $A + $B * 0.1' "$A" "$B"
# Python 2
python -c 'import sys ; a = float(sys.argv[1]) ; b = float(sys.argv[2]) ; print a+b*0.1' "$A" "$B"
# Python 3
python3 -c 'import sys ; a = float(sys.argv[1]) ; b = float(sys.argv[2]) ; print(a+b*0.1)' "$A" "$B"
# BC
bc <<< "scale=1 ; $A + $B * 0.1"

Answer 18

As others have indicated, bash does not have built-in floating-point operators.
You can implement floating-point in bash, even without using calculator programs like bc and awk, or any external programs for that matter.
I'm doing exactly this in my project, shellmath, in three basic steps:
Break the numbers down into their integer and fractional parts
Use the built-in integer operators to process the parts separately while being careful about place-value and carrying
Recombine the results
As a teaser, I've added a demo script that calculates e using its Taylor series centered at x=0.
Please check it out if you have a moment. I welcome your feedback!

Answer 19

Bash can calculate floating point results just fine without any other programs.
Bash independently even can calculate π to the ninth decimal place accurately.
Example:
calc=104348/33215
accuracy=9
calc99p9=$((10**$accuracy))*$calc
result99p9=$((calc99p9))
result=${result99p9: -${#result99p9}: -$accuracy}.${result99p9: -$accuracy}
echo Bash calculated pi to be $result
results in
Bash calculated pi to be 3.141592653

Answer 20

Dividend = Divisor × Quotient + Remainder
Lets just calculate the quotient and remainder.
As well as concatenate those strings in a variable.
New way valid only to log_decimal divisors:
function main() {
bar=10030
divisor=100
# divisor=50
quotient=$((bar / divisor))
# remainder=$((bar - v_int * divisor))
remainder=$((bar % divisor))
remainder_init=$remainder
printf "%-15s --> %s\n" "quotient" "$quotient"
printf "%-15s --> %s\n" "remainder" "$remainder"
cnt=0
while :; do
remainder=$((remainder * 10))
aux=$((remainder / divisor))
printf "%-15s --> %s\n" "aux" "$aux"
[[ aux -ne 0 ]] && break
((cnt += 1))
printf "%-15s --> %s\n" "remainder" "$remainder"
done
printf "%-15s --> %s\n" "cnt" "$cnt"
printf "%-15s --> %s\n" "aux" "$aux"
printf $quotient
printf "."
for i in $(seq 1 $cnt); do printf "0"; done
printf $remainder_init
}
clear
main
Old wrong way:
bar=1234 \
&& divisor=1000 \
&& foo=$(printf "%s.%s" $(( bar / divisor )) $(( bar % divisor ))) \
&& printf "bar is %d miliseconds or %s seconds\n" $bar $foo
Output: bar is 1234 miliseconds or 1.234 seconds

Answer 21

here is awk command: -F = field separator == +
echo "2.1+3.1" | awk -F "+" '{print ($1+$2)}'

Answer 22

how precise do you need the output to be ? if an approximation via binning is already acceptable for your use case, you can even go one step further and take advantage of POSIX exit codes [0:256) (all other integers mod'd back to that range).
eg : in gawk/nawk/mawk-1, it already gives me epoch seconds down to the integer level, but I wanted to expand that to extract near-milliseconds precision but not overly pedantic about it, i run this command in a POSIX shell
exit $(( 10#` gdate +%5N ` * 256 / 100000 ))
directly assigning a 5-digit integer representing 0.xxxxxx output of gnu-date into 1 of 256 bins, then un-doing that 256 once awk obtains the exit code of the system() call, i.e. the selected bin #. I found this approach to have lower overhead than using a full getline call.
This method also directly captures the output into the POSIX exit code instead of also having an extra terminal print out.
(the shell arithmetic auto floors it to integer if written this way instead of * 0.0256). Putting together into an awk function, it would resemble this. The 10# is to force base-10 to prevent posix shell interpreting "01733" as an octal number.
function msecs() { # n x 2**-8 = n divided by 256
return 2^-8 * \
system( "exit \44\50\50 " \
" 10\43\140 gdate \53" \
"%5N\140 \52 " \
"256 \57 100000 \51\51" )
}
for my own code, i apply another 0.6% haircut to account for shell-overhead.

Answer 23

Let us take an example to undertsand, if u want to find average of n array elements (Ofcourse the average will be in float/decimals)
declare -a arr
echo "How many numbers you want to enter?"
read n
echo "Enter the Array Elements"
for(( i=0 ; i<$n ; i++))
do
read array_elements
arr[$i]="$array_elements"
done
sum=0
for i in "${arr[#]}"
do
#sum and avg
sum=$(($sum + $i))
#average will come in decimals
avg=`echo $sum / $n | bc -l`
done
# Output results:
printf "Average of Array Elements %.2f:" $avg
So we will use "| bc -l" to do floating calculations