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Groups > gnu.bash.bug > #15789
| From | Chet Ramey <chet.ramey@case.edu> |
|---|---|
| Newsgroups | gnu.bash.bug |
| Subject | Re: bash problem |
| Date | 2020-01-03 08:34 -0500 |
| Organization | ITS, Case Western Reserve University |
| Message-ID | <mailman.1921.1578058467.1979.bug-bash@gnu.org> (permalink) |
| References | <LOYP265MB1902BD6625252A65C2F6A5D88D200@LOYP265MB1902.GBRP265.PROD.OUTLOOK.COM> <fce0b821-0246-ca58-578a-5ffe7dc54124@case.edu> |
On 1/2/20 8:45 AM, Wheatley, Martin R wrote: > It appears that there may be an issue when bash terminates a shell as a result of TMOUT expiring. > > I was attempting to enter "dc_command show media | grep YD" to the shell and this is the terminal output... > > OLDuser@hostname $ dc_command show mediatimed out waiting for input: auto-logout > NEWuser@hostname $ | grep YD > -bash: syntax error near unexpected token `|' I don't know what you have TMOUT set to, but it seems kind of small from that example. > If the command line had been something like " cd fred && rm -rf *" > and the TMOUT had expired just after the '&&' had been entered then the "rm-rf *" would have been executed in the `cwd` of the parent process rather than in 'fred' Yes. Since a child and its parent process share file descriptors (and file pointers), terminal input not consumed by the child is available for the parent. > > It looks like the TMOUT isn't re-started after any shell input but only after a '\n' is seen. That's how it's documented: TMOUT is the number of seconds to wait for "a line of input after issuing the primary prompt." It issues the prompt and waits for a complete line of input. That way it can work whether or not you're using readline. -- ``The lyf so short, the craft so long to lerne.'' - Chaucer ``Ars longa, vita brevis'' - Hippocrates Chet Ramey, UTech, CWRU chet@case.edu http://tiswww.cwru.edu/~chet/
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Re: bash problem Chet Ramey <chet.ramey@case.edu> - 2020-01-03 08:34 -0500
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