On 06/05/2012 11:35 AM, Dan Nelson wrote:
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:
Ah, OK, that makes sense, thanks...
--
-----------------------------------------------------------------------
Tim Daneliuk
_______________________________________________
[email protected] mailing list
http://lists.freebsd.org/mailman/listinfo/freebsd-questions
To unsubscribe, send any mail to "[email protected]"
