Where it worked...
tt=0000005
echo $((tt))
tt=$(($tt+1))
echo $((tt))
5
6
Somewhere it stopped working...
tt=0056505
echo $((tt))
tt=$(($tt+1))
echo $((tt))
23877
23878
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0
Elrik
- 25
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1bash interprets numbers that start with "0" as octal, rather than decimal, and 0056505 (octal) = 23877 (decimal). See ["How can I do bash arithmetic with variables that are numbers with leading zeroes?"](https://stackoverflow.com/questions/53075017/how-can-i-do-bash-arithmetic-with-variables-that-are-numbers-with-leading-zeroes) and ["How can I increment a number in a while-loop while preserving leading zeroes (BASH < V4)"](https://stackoverflow.com/questions/21592667/how-can-i-increment-a-number-in-a-while-loop-while-preserving-leading-zeroes-ba). – Gordon Davisson Aug 10 '22 at 03:07
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1BTW, for incrementing the variable, a simpler way to write it is `((tt++))`. – user1934428 Aug 10 '22 at 06:50
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@user1934428 `((tt++))` won't work if the initial value of `tt` is, for instance, `08`. – M. Nejat Aydin Aug 10 '22 at 06:56
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@MNejatAydin: Because `08` is not a number, as explained by Gordon Davisson in his comment above. If you initialize it to `8`, it works. BTW, if you initialize the variable to something like `foobar`, it wouldn't work either. – user1934428 Aug 10 '22 at 07:00
1 Answers
0
You need to trim the leading zeros first
$ tt=0056505
$ tt=$(echo $tt | sed 's/^[0]*//g')
$ echo $((tt))
56505
$ tt=$(($tt+1))
$ echo $((tt))
56506
WeDBA
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1You can also say `tt=$(( 10#$tt + 1 ))` to tell bash to interpret `$tt` as a decimal number. – tshiono Aug 10 '22 at 03:18
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Creating a subshell and calling an external program to trim leading `0`s from a shell variable is superfluous. Trimming could be done via shell parameter expansions: `tt=${tt#"${tt%%[^0]*}"}` (instead of `tt=$(echo $tt | sed 's/^[0]*//g')`). Or, in bash, `tt=${tt##+(0)}` after setting the `extglob` shell option by `shopt -s extglob`. – M. Nejat Aydin Aug 10 '22 at 06:48