How do I strip leading zeroes for math in a shell script?

Not to be confused with this Dave Taylor, I'm the Dave Taylor from the game industry, and I'm visiting because my uncle Paul Taylor thought I might have been doing this as I share your penchant for answering things.

It's a neat site, and I think you're providing a darned handy service, btw.

I think there's a small bug with both solutions. You've got an issue if one of the lines reads thusly:

0000000000

You'd want to strip that down to just "0", but both algorithms you use basically delete the line, which could cause some pretty serious issues, depending on what's eating the output.

Darn, you're right, Dave! Using the 'sed' solution, simply add a test after the conditional:

if [ -z "$valueWithoutZeroes" ] ; then
valueWithoutZeroes="0"
fi

that should do the trick!

I figure out a much simpler way of doing it:

newValue=`expr $oldValue + 0`

Note that this only works with "data that are amounts", as the question says.

Here's my usual solution, pretty much equivalent to Alejandro's use of expr:
newValue=`echo $oldValue | awk '{print $1 + 0}'`
or
newValue=`echo $oldValue | awk '{printf "%d\n", $1}'

I'm fairly new to this and I do follow the logic. However, I have problems trying to tie it all up together. For instance I have a file 'benshrs' with figures I need to add up in columns 21 thru 27. Some are all zeroes. What's the best way (complete script)that combines a. the cut -c 21-27 b. strip the leading zeroes except when it is all zeroes, c. add up the figures in do ... done loop, and lastly print the final sum.

My crude attempt (below) has failed miserably!
hrs_total=0.00
for line_num in `cut -c 21-27 $0`
do
line_hrs= `expr `sed 's/0*//' $line_num` + 0`
(hrs_total=$hrs_total + $line_hrs)
done
hrs_total=`echo "scale=2;$hrs_total/100" | bc`
echo "Hours Total = $hrs_total"

I'm not surprised this isn't working, Ben: you can't use floating point / real numbers in a shell script, so even your initial assignment of hrs_total=0.00 is going to fail before you even get into the for loop. It's possible that you need to do this in awk or, better, Perl, to get it working.

Not that your option wouldn't work, Dave, but I'm curious if it isn't a bit of an overkill solution. Why not simply employ something like:

$ typeset -i example1="00001"
$ typeset -i example2="a0001"
$ typeset -i example3="00000"
$ echo $example1 + 10 |bc
11
$ echo $example2 + 10 |bc
10
$ echo $example3 + 10 |bc
10
$ print $example1
1
$ print $example2
0
$ print $example3
0
===

This approach is available to both ksh and bash, and would immediately strip out the zeroes or otherwise negate the value to "0" if there were some alpha values. Same applies to a fully zero-filled value. This eliminates the need for any additional function definitions at all...but is this a matter for portability? Sure it restricts the math to integers, but sed would totally skew the numbers if it was used to massage the value.

[Your page doesn't seem to restrict the discussion to a particular shell that I can see.]

Cool. Never knew about the "typeset" built-in to the shell. Very interesting solution, thanks!

The typeset -i (or declare -i in more recent versions of bash) doesn't work if the value is greater than 7 -- it's interpreted as an octal number.

$ declare -i x=009
bash: declare: 009: value too great for base (error token is 009)
$ declare -i x=010
$ echo $x
8

The most reliable solution so far is using expr + 0 I think.