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How to increment a numeric variable in bash? I tried with var=$var+1 and var=($var+1) but it doesn't work. My variable is a number, but bash think that is a string. My bash version is 4.2.45(1)-release (x86_64-pc-linux-gnu) and I use Ubuntu 13.10.

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up vote 283 down vote accepted

There are more than one way to increment a variable in bash, but what you tried is not correct.

You can use for example arithmetic expansion:


or only:




or even:


Or you can use let:

let "var=var+1"

or only:

let "var+=1"

or even:

let "var++"

See also:

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or ((++var)) or ((var=var+1)) or ((var+=1)). – gniourf_gniourf Dec 3 '13 at 18:30
or var=$(expr $var + 1) – chilicuil Dec 3 '13 at 23:20
Curiously, var=0; ((var++)) returns an error code while var=0; ((var++)); ((var++)) does not. Any idea why? – phunehehe Mar 26 '14 at 9:36
@RaduRădeanu I'm sure, the same thing happens in zsh too. Actually maybe I didn't make myself clear. This prints 1 (an error code) instead of 0 (successful): var=0; ((var++)); echo $?. – phunehehe Mar 27 '14 at 11:08
@phunehehe Look at help '(('. The last line says: Returns 1 if EXPRESSION evaluates to 0; returns 0 otherwise. – Radu Rădeanu Mar 27 '14 at 11:41
var=$((var + 1))

Arithmetic in bash uses $((...)) syntax.

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Vastly better than the accepted answer. In just 10% as much space, you managed to provide enough examples (one is plenty - nine is overkill to the point when you're just showing off), and you provided us with enough info to know that ((...)) is the key to using arithmetic in bash. I didn't realize that just looking at the accepted answer - I thought there was a weird set of rules about order of operations or something leading to all of the parenthesis in the accepted answer. – ArtOfWarfare Jun 22 at 13:56

Performance Analysis of various options

Thanks to Radu Rădeanu answer that provides the following list of ways to increment a variable in bash.

  • var=$((var+1))
  • ((var=var+1))
  • ((var+=1))
  • ((var++))
  • let "var=var+1"
  • let "var+=1" 
  • let "var++"

Having so many options leads to these two questions:

  1. Is there a performance difference between them?
  2. If so which, which performs best?

Incremental performance test code:


# To focus exclusively on the performance of each type of increment
# statement, we should exclude bash performing while loops from the
# performance measure. So, lets time 7 individual scripts that
# increment $i in their own unique way. 
until [ $x = "10000000" ]; do
    echo 'i=$((i+1))' >> t0
    echo '((i=i+1))' >> t1
    echo '((i+=1))' >> t2
    echo '((i++))' >> t3
    echo 'let "i=i+1"'>> t4
    echo 'let "i+=1"' >> t5
    echo 'let "i++"' >> t6

chmod +x t0 t1 t2 t3 t4 t5 t6

echo -n 'i=$((i+1)) '; { time ./t0 ;} |& grep user
echo -n '((i=i+1))  '; { time ./t1 ;} |& grep user
echo -n '((i+=1))   '; { time ./t2 ;} |& grep user
echo -n '((i++))    '; { time ./t3 ;}|& grep user
echo -n 'let "i=i+1"    '; { time ./t4 ;} |& grep user
echo -n 'let "i+=1" '; { time ./t5 ;} |& grep user
echo -n 'let "i++"  '; { time ./t6 ;} |& grep user
rm t0 t1 t2 t3 t4 t5 t6


i=$((i+1))  user    1m20.797s
((i=i+1))   user    1m3.912s
((i+=1))    user    1m0.430s
((i++))     user    0m53.694s
let "i=i+1" user    1m49.539s
let "i+=1"  user    1m46.535s
let "i++"   user    1m40.828s

Looking through the comments, one can also find several other ways to increment a variable in bash. ((++i)) performed very similar to ((i++)) and i=(expr $i +1) performed similar to let statements


It seems bash is fastest at performing ((i++)). let and expr statements seem particularly slow.

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There's also this:

var=`expr $var + 1`

Take careful note of the spaces and also ` is not '

While Radu's answers, and the comments, are exhaustive and very helpful, they are bash-specific. I know you did specifically ask about bash, but I thought I'd pipe in since I found this question when I was looking to do the same thing using sh in busybox under uCLinux. This portable beyond bash.

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There's one method missing in all the answers - bc

$ VAR=7                                                                   

$ bc <<< "$VAR+2"                                                         

$ echo $VAR                                                               

$ VAR=$( bc <<< "$VAR+1" )

$ echo $VAR                                                               
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