Date: Tue, 6 Nov 2001 09:20:01 -0800 (PST) From: Thomas Quinot <thomas@cuivre.fr.eu.org> To: freebsd-bugs@FreeBSD.org Subject: Re: bin/31627 sh(1) is broken - loss of data! Message-ID: <200111061720.fA6HK1O58624@freefall.freebsd.org>
next in thread | raw e-mail | index | archive | help
The following reply was made to PR bin/31627; it has been noted by GNATS. From: Thomas Quinot <thomas@cuivre.fr.eu.org> To: Eugene Grosbein <eugen@svzserv.kemerovo.su> Cc: stable@freebsd.org, freebsd-gnats-submit@freebsd.org Subject: Re: bin/31627 sh(1) is broken - loss of data! Date: Tue, 6 Nov 2001 18:18:34 +0100 Le 2001-11-06, Eugene Grosbein écrivait : > #!/bin/sh > string=`printf "\21"` > echo $string | hd > Replace 21 with 201 and rerun. You see: > 00000000 0a |.| > 00000001 Can't reproduce here for the value \201, but for the other values you mention it looks like perfectly normal and expected behaviour from sh(1). It is not surprising at all that some characters "disappear" here: since $string appears unquoted, any character which is whitespace w.r.t. shell parsing rules won't be passed to echo. Try to quote your string: echo "$string" | hd In your other example, you use the 'read' builtin to get characters from jot, but read is /also/ defined to apply shell field splitting rules. A correct version of your test follows: #!/bin/sh -x for n in `jot 256 0` do c="`jot -c 1 $n`" echo "$c" | wc -c | grep -v 2 && echo "$n" done which correctly produces the following output: 1 0 1 10 because a shell variable cannot contain a null character (which is a string end marker), and backquote expansion is defined to remove trailing newlines. This is legal and expected behaviour, not a bug. Thomas. -- Thomas.Quinot@Cuivre.FR.EU.ORG To Unsubscribe: send mail to majordomo@FreeBSD.org with "unsubscribe freebsd-bugs" in the body of the message
Want to link to this message? Use this URL: <https://mail-archive.FreeBSD.org/cgi/mid.cgi?200111061720.fA6HK1O58624>