AW: Questions to bash "read" builtin functionality

[Fiedler Roman]

> -Ursprüngliche Nachricht-
> Von: Chet Ramey [mailto:chet.ra...@case.edu]
> Gesendet: Samstag, 15. Dezember 2012 23:53
> 
> On 12/14/12 6:28 AM, Fiedler Roman wrote:
> > Hello list,
> >
> > One of our bash-scrips failed with very low frequency but randomly. The
> result was that exactly 1 byte was lost, so the string returned by "read -t 1"
> was too short. The culprit seems to be the built-in read function itself, the
> probability of failure was about 1:10 in our case.
> >
> > After analysis and creation of a reliable reproducer, I'm not sure if this 
> > is a
> programming error in bash or a bug in our script as the result of suboptimal
> documentation.
> >
> > * Description:
> >
> > When "read" builtin is used with timeout setting, both with "-t" option or 
> > by
> setting the "TMOUT" environment variable, and the timeout occurs when
> some bytes of line were already received by bash, then those bytes are
> silently dropped. The next "read" returns the remainder of the line.
> 
> You don't say what version of bash you're using, ...

Sorry, forgot to mention it:

# bash --version
GNU bash, version 4.2.24(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2011 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later 

> but the following change  went in with bash-4.0:
>
> k.  Changed the behavior of the read builtin to save any partial input 
> received
> in the specified variable when the read builtin times out.  This also
> results in variables specified as arguments to read to be set to the empty
> string when there is no input available.  When the read builtin times out,
> it returns an exit status greater than 128.

That is strange: If I understand correctly, following script combined with the 
one from first mail should still fail, but with different error. But it fails 
in a way similar to discarding partial input, not saving it. Could it be, that 
Ubuntu-bash works differently? Output on my  side is:

Script:

#!/bin/bash
line=""
while true; do
  read -t 1 line
  status="$?"
  if [ "${status}" != "0" ]; then
echo "Read status ${status}, value \"${line}\"" >&2
continue
  fi
  if [ "${line}" != "Status: OK" ]; then
echo "FAILED READ: \"${line}\"" >&2
exit 1
  fi
done

# ./FragmentedSend.py | ./BashReadTest 
Read status 142, value ""
Read status 142, value ""
FAILED READ: "us: OK"
Traceback (most recent call last):
  File "./FragmentedSend.py", line 16, in 
os.write(1, nextSendData[0:sendLength])
OSError: [Errno 32] Broken pipe

>From your description, I would expect 

--- HYPOTHETICAL-OUTPUT---
# ./FragmentedSend.py | ./BashReadTest 
Read status 142, value ""
Read status 142, value "Stat"  << the partial read
FAILED READ: "us: OK"
Traceback (most recent call last):
  File "./FragmentedSend.py", line 16, in 
os.write(1, nextSendData[0:sendLength])
OSError: [Errno 32] Broken pipe
--- HYPOTHETICAL-OUTPUT---

Is this consistent with what you would be expecting!

> That is not clearly stated in the manual's description of `read'.

Perhaps it could be added, at least "see specification [URL] for details on 
read input handling"

Roman

Re: AW: Questions to bash "read" builtin functionality

[Chet Ramey]

On 12/17/12 3:34 AM, Fiedler Roman wrote:

> That is strange: If I understand correctly, following script combined with 
> the one from first mail should still fail, but with different error. But it 
> fails in a way similar to discarding partial input, not saving it. Could it 
> be, that Ubuntu-bash works differently? Output on my  side is:
> 
> Script:
> 
> #!/bin/bash
> line=""
> while true; do
>   read -t 1 line
>   status="$?"
>   if [ "${status}" != "0" ]; then
> echo "Read status ${status}, value \"${line}\"" >&2
> continue
>   fi
>   if [ "${line}" != "Status: OK" ]; then
> echo "FAILED READ: \"${line}\"" >&2
> exit 1
>   fi
> done
> 
> # ./FragmentedSend.py | ./BashReadTest

What is FragmentedSend.py?  If I can use the same scripts you are, I can
try to reproduce it.

Chet
-- 
``The lyf so short, the craft so long to lerne.'' - Chaucer
 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRUc...@case.eduhttp://cnswww.cns.cwru.edu/~chet/

AW: AW: Questions to bash "read" builtin functionality

[Fiedler Roman]

> Von: Chet Ramey [mailto:chet.ra...@case.edu]
> Gesendet: Montag, 17. Dezember 2012 15:10
> 
> On 12/17/12 3:34 AM, Fiedler Roman wrote: 
> > That is strange: If I understand correctly, following script combined with 
> > the
> one from first mail should still fail, but with different error. But it fails 
> in a way
> similar to discarding partial input, not saving it. Could it be, that 
> Ubuntu-bash
> works differently? Output on my  side is:
> >
> > Script:
> >
> > #!/bin/bash
> > line=""
> > while true; do
> >   read -t 1 line
> >   status="$?"
> >   if [ "${status}" != "0" ]; then
> > echo "Read status ${status}, value \"${line}\"" >&2
> > continue
> >   fi
> >   if [ "${line}" != "Status: OK" ]; then
> > echo "FAILED READ: \"${line}\"" >&2
> > exit 1
> >   fi
> > done
> >
> > # ./FragmentedSend.py | ./BashReadTest
> 
> What is FragmentedSend.py?  If I can use the same scripts you are, I can
> try to reproduce it.

Sure, it was in the first mail of  this thread:

base64 -d <