[SOLVED] Numeric addition in bash

garion · 09-06-2012, 07:29 AM

I am writing a script that will retry an action a number of times before it fails and I discovered both of these work:

RETRY=$(( $RETRY + 1 ))
RETRY=$(( RETRY + 1 ))

It also seems that they behave the exact same way. The first example makes sense, but not the second. I googled, but I could not find an explanation for how or why the second would work.

Any info would be appreaciated.

pan64 · 09-06-2012, 07:35 AM

see here (for example):
http://www.bashguru.com/2010/12/math...l-scripts.html
it works: bash knows RETRY cannot be a number, so it is a variable.

garion · 09-06-2012, 07:46 AM

So bash is simply smart enough to know what it is supposed to be. Cool! Thank you very much for the quick response.

And screw those windows-weanies who claim there is no support for Linux!

grail · 09-06-2012, 08:12 AM

You can also do:

Code:

((RETRY++))

David the H. · 09-07-2012, 01:53 AM

It's not just bash. All posix-based shells should work in basically the same way. The ((..)) construct is only available in bash/ksh/zsh, however.

See here for much more on shell arithmetic: arithmetic expressions

Edit:

Note that the shell is only smart enough to determine if a variable contains a numeral or not. For other operations you need to use the "$".

Code:

#build an entire expression from variables
$ a=4 b=+ c=3
$ echo $(( a b c ))
bash: +: syntax error: operand expected (error token is "+")
$ echo $(( a $b c ))
7

#add the total number of array entries to the number in the last entry
$ array=( 1 2 3 )
$ echo "$(( array[2] + ${#array[@]} ))"
6