I have tried to increment a numeric variable using both var=$var+1 and var=($var+1) without success. The variable is a number, though bash appears to be reading it as a string.

Bash version 4.2.45(1)-release (x86_64-pc-linux-gnu) on Ubuntu 13.10.

0

9 Answers

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

You can use, for example, arithmetic expansion:

var=$((var+1)) ((var=var+1)) ((var+=1)) ((var++))

Or you can use let:

let "var=var+1" let "var+=1" let "var++"

See also: .

17
var=$((var + 1))

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

1

Various options to increment by 1, and performance analysis

Thanks to Radu Rădeanu's answer that provides the following 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++"

There are other ways too. For example, look in the other answers on this question.

let var++ var=$((var++)) ((++var)) { declare -i var var=var+1 var+=1 } { i=0 i=$(expr $i + 1) }

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:

#!/bin/bash # To focus exclusively on the performance of each type of increment # statement, we should exclude bash performing while loops from the # performance measure. So, let's time individual scripts that # increment $i in their own unique way. # Declare i as an integer for tests 12 and 13. echo > t12 'declare -i i; i=i+1' echo > t13 'declare -i i; i+=1' # Set i for test 14. echo > t14 'i=0; i=$(expr $i + 1)' x=100000 while ((x--)); do echo >> t0 'i=$((i+1))' echo >> t1 'i=$((i++))' echo >> t2 '((i=i+1))' echo >> t3 '((i+=1))' echo >> t4 '((i++))' echo >> t5 '((++i))' echo >> t6 'let "i=i+1"' echo >> t7 'let "i+=1"' echo >> t8 'let "i++"' echo >> t9 'let i=i+1' echo >> t10 'let i+=1' echo >> t11 'let i++' echo >> t12 'i=i+1' echo >> t13 'i+=1' echo >> t14 'i=$(expr $i + 1)' done for script in t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10 t11 t12 t13 t14; do line1="$(head -1 "$script")" printf "%-24s" "$line1" { time bash "$script"; } |& grep user # Since stderr is being piped to grep above, this will confirm # there are no errors from running the command: eval "$line1" rm "$script" done

Results:

i=$((i+1)) user 0m0.992s i=$((i++)) user 0m0.964s ((i=i+1)) user 0m0.760s ((i+=1)) user 0m0.700s ((i++)) user 0m0.644s ((++i)) user 0m0.556s let "i=i+1" user 0m1.116s let "i+=1" user 0m1.100s let "i++" user 0m1.008s let i=i+1 user 0m0.952s let i+=1 user 0m1.040s let i++ user 0m0.820s declare -i i; i=i+1 user 0m0.528s declare -i i; i+=1 user 0m0.492s i=0; i=$(expr $i + 1) user 0m5.464s

Conclusion:

It seems bash is fastest at performing i+=1 when $i is declared as an integer. let statements seem particularly slow, and expr is by far the slowest because it is not a built into bash.

6

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.

2

If you declare $var as an integer, then what you tried the first time will actually work:

$ declare -i var=5 $ echo $var 5 $ var=$var+1 $ echo $var 6

Reference: Types of variables, Bash Guide for Beginners

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

$ VAR=7 $ bc <<< "$VAR+2" 9 $ echo $VAR 7 $ VAR=$( bc <<< "$VAR+1" ) $ echo $VAR 8

bc is specified by POSIX standard, so should be present on all versions of Ubuntu and POSIX-compliant systems. The <<< redirection could be altered to echo "$VAR" | bc for portability, but since the question asks about bash - it's OK to just use <<<.

The return code 1 issue is present for all default variants (let, (()), etc.). This often causes trouble, e.g., in scripts that use set -o errexit. Here is what I am using to prevent error code 1 from math expressions that evaluate to 0;

math() { (( "$@" )) || true; } math a = 10, b = 10 math a++, b+=2 math c = a + b math mod = c % 20 echo $a $b $c $mod #11 12 23 3

This has to be the worst way to accomplish such a simple task but I just wanted to document it for fun I guess (complete opposite of code golf).

$ var=0 $ echo $var 0 $ var="$(python -c 'print('$var'+1)')" $ echo $var 1

or

$ var="$(printf '%s\n' $var'+1' | bc)" $ echo $var 1

Seriously use one of the other much better choices here.

This is the safe bet

(( var = var + 1 ))

If the resulting value is non zero, then setting exit on error will stop Your script

set -e var=0 (( var++ )) echo You will never get here