Date: Tue, 5 Jun 2012 11:35:22 -0500 From: Dan Nelson <dnelson@allantgroup.com> To: Tim Daneliuk <tundra@tundraware.com> Cc: FreeBSD Mailing List <freebsd-questions@freebsd.org> Subject: Re: Possible /bin/sh Bug? Message-ID: <20120605163522.GD16874@dan.emsphone.com> In-Reply-To: <4FCE287D.3090501@tundraware.com> References: <4FCE287D.3090501@tundraware.com>
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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 dnelson@allantgroup.com
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