Possible /bin/sh Bug?
Randy Pratt
bsd-unix at embarqmail.com
Tue Jun 5 17:32:43 UTC 2012
On Tue, 05 Jun 2012 10:40:45 -0500
Tim Daneliuk <tundra at tundraware.com> wrote:
> Given this script:
> #!/bin/sh
>
> foo=""
> while read line
> do
> foo="$foo -e"
> done
> echo $foo
>
> Say I respond 3 times, I'd expect to see:
>
> -e -e -e
>
> Instead, I get:
>
> -e -e
The last line "echo $foo" is what is getting confused. At the end of
3 passes, $foo contains " -e -e -e" so when the last line is executed,
it looks like:
echo -e -e -e
The first -e is probably being interperted by "echo" as a flag
( echo -e ) and then only prints the last two -e.
Its easier to see if you execute the script with xtrace:
sh -x /path/to/script
I'd recommend that you write the last line with quotes:
echo "$foo"
and I think it'll produce the results you expect.
HTH,
Randy
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