Ok, that makes sense, but why doesn't it work if I change the delimiter? read -t 3 -d q *press random keys without pressing q*
I think the same should happen here, I'm asking bash to read as much input as it can until it reads a q. Since I don't press q, whatever I typed should be used as typeahead. But that's not the case, and I don't understand why. 2015-01-04 17:21 GMT+01:00, Chet Ramey <chet.ra...@case.edu>: > On 1/4/15 12:45 AM, isabella parakiss wrote: >> I'm trying to use read -t in an interactive shell >> >> read -t 3 *press random keys* >> >> Everything i press is now part of the next command in the prompt. >> It only happens when the delimiter is a \n. >> Is this intended? What's the point? > > It's difficult for me to tell what the question is here. You've asked > for bash to read as much input as it can until it reads a newline, with > the read aborted if you don't press a newline within three seconds. Since > you don't press newline, read(2) doesn't return anything and whatever > you've typed is left in the input buffer as typeahead for readline. When > readline is called, it is able to read all of the typeahead and use it as > part of the next input line. > > -- > ``The lyf so short, the craft so long to lerne.'' - Chaucer > ``Ars longa, vita brevis'' - Hippocrates > Chet Ramey, ITS, CWRU c...@case.edu http://cnswww.cns.cwru.edu/~chet/ >
