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I have a simple script that deals with hours and minutes.

If I want to calculate number of minutes since midnight having a string s hh:mm I tried splitting string then doing hh * 60 + mm

My problem is, while

$ (( tot = 12 * 60 + 30 ))
$ echo $tot
750

instead

$ (( tot = 09 * 60 + 30 ))
bash: ((: tot = 09: value too great for base (error token is "09")

As far as I understand string 09 is not to be intended as a base 10 number.

Is there a better way than simply removing leading zeros in string?

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2 Answers 2

up vote 10 down vote accepted
h=09; m=30;(( tot = 10#$h * 60 + 10#$m )); echo $tot  

The number before the # is the radix (or base)
The number after the # must be valid for the radix
The output is always decimal
You can use a radix of 2 thru 64 (in GNU bash 4.1.5)

As noted by enzoyib, the old alternative of $[expression] is depricated, so it is beter to use the POSIX compliant $((expr))

$(( 2#1)) ==  1
$((16#F)) == 15
$((36#Z)) == 35  

I'm not sure which 'digits' are used after Z

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what about a litte more information about $[10#09] ??? –  neurino Aug 24 '11 at 8:53
    
I enclosed it in hh extraction: hh=$[10#${hhmm%:*}] the hell of readability! :D Thanks for the quick answer! –  neurino Aug 24 '11 at 9:07
3  
+1, but from bash's manual page: "The old format $[expression] is deprecated and will be removed in upcoming versions of bash.". Better to use $((expr)) that is POSIX compliant. –  enzotib Aug 24 '11 at 11:01
1  
The ksh and bash manpages describe the digits for bases higher than 36: 0-9, a-z, A-Z, @, _ (bases 36 and lower are case insensitive). –  Chris Johnsen Aug 25 '11 at 6:35

A leading zero on a numeric constant in shell arithmetic expressions denotes an octal constant.

Here's a portable way of deleting initial zeros:

h=${h#${h%%[!0]*}}; [ -n "$h" ] || h=0

In bash, ksh or zsh, you can explicitly specify base 10 with $((10#$h)).

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