In the last episode (Jun 05), Tim Daneliuk said:
> 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
>
> Linux appears to do the right thing here, so this seems like it
> is a bug ... or am I missing something?
echo takes a -e flag, so it eats the first one. Bash does the same thing,
so any Linux that uses bash as /bin/sh will also. You must be testing on a
Linux that uses something else as /bin/sh. Better to use the printf command
if you are worried about compatibility.
echo [-e | -n] [string ...]
Print a space-separated list of the arguments to the standard
output and append a newline character.
-n Suppress the output of the trailing newline.
-e Process C-style backslash escape sequences. The echo
command understands the following character escapes:
--
Dan Nelson
dnel...@allantgroup.com
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