var=2 n=3 var+=$n
In bash/ksh, use an (( arithmetic context ))
(( arithmetic context ))
(( var += n ))
or declare the variable as an integer type:
declare -i var=2 n=4 var+=$n
For POSIX sh, use an $((arithmetic expansion)):
$((arithmetic expansion))
var=$((var+n))
The problematic code attempts to add 2 and 3 to get 5.
Instead, += on a string variable will concatenate, so the result is 23.
+=
If you do want to concatenate a number, for example to append trailing zeroes, you can silence the warning by quoting the number:
var+="000"
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