How to do exponentiation in Bash

Question

I tried

echo 10**2

which prints 10**2. How to calculate the right result, 100?

Solution 1

You can use the let builtin:

let var=10**2   # sets var to 100.
echo $var       # prints 100

or arithmetic expansion:

var=$((10**2))  # sets var to 100.

Arithmetic expansion has the advantage of allowing you to do shell arithmetic and then just use the expression without storing it in a variable:

echo $((10**2)) # prints 100.

For large numbers you might want to use the exponentiation operator of the external command bc as:

bash:$ echo 2^100 | bc
1267650600228229401496703205376

If you want to store the above result in a variable you can use command substitution either via the $() syntax:

var=$(echo 2^100 | bc)

or the older backtick syntax:

var=`echo 2^100 | bc`

Note that command substitution is not the same as arithmetic expansion:

$(( )) # arithmetic expansion
$( )   # command substitution

Solution 2

Various ways:

Bash

echo $((10**2))

Awk

awk 'BEGIN{print 10^2}'  # POSIX standard
awk 'BEGIN{print 10**2}' # GNU awk extension

bc

echo '10 ^ 2' | bc

dc

dc -e '10 2 ^ p'

Solution 3

Actually var=$((echo 2^100 | bc)) doesn't work - bash is trying to do math inside (()). But a command line sequence is there instead so it creates an error

var=$(echo 2^100 | bc) works as the value is the result of the command line executing inside ()

Comments

[DarkDust]

Instead of $(( ... )) it's also possible to use $[ ... ]. I find the later visually more appealing but that's just my taste. Don't know if there are any differences between the two, though. It seems they behave the same.

[slezica]

Alberto: all this parenthesis, lets and $ are needed because bash, being a simple command-line interpreter, does better assuming everything is a string and should be treated as such, unless explicitly indicated otherwise. Saludos!

[wisbucky]

@DarkDust $[...] is an older deprecated syntax. unix.stackexchange.com/questions/209833/…

[bballdave025]

This is very helpful You should put it as a comment to the answer above.

[bballdave025]

Good catch, by the way; the other, accepted commands followed the $((...)) pattern (Double Parentheses Construct, source Advanced Bash-Scripting Guide, "permits arithmetic expansion and evaluation"). Here, as you've noted, we need Command Substitution, i.e. the $(...) construct. The ABS guide notes, "The $(...) has superceded backticks". The Bash Beginners Guide, section 3.4.5-6, has both described right next to each other.

[Wrichik Basu]

NOTE: $(( ... )) does not work. Instead, $( ... ) has to be used. See @firefly's answer below.

[Nathaniel M. Beaver]

@WrichikBasu I submitted an edit; it's fixed now.