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jupyter-with-timeout: Don't spin wait so much

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`accept-process-output` will return when any output has been read from a
process (after it has been handled by the process filter function) which is
usually the time when we want to check on the wait condition so there is no
need to have such a low time resolution.
This commit is contained in:
Nathaniel Nicandro 2019-03-17 02:05:00 -05:00
parent c0d13396bb
commit cdc826259e
1 changed files with 7 additions and 3 deletions
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10
jupyter-base.el
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@ -222,14 +222,18 @@ SPEC takes the form (PROGRESS SECONDS TIMEOUT-FORMS...).
(declare (indent 1) (debug ((form form body) body)))
(let ((res (make-symbol "res"))
(prog (make-symbol "prog"))
(prog-msg (make-symbol "prog-msg")))
(prog-msg (make-symbol "prog-msg"))
(timeout (make-symbol "timeout"))
(wait-time (make-symbol "wait-time")))
`(let* ((,res nil)
(,prog-msg ,(pop spec))
(,timeout ,(pop spec))
(,wait-time (/ ,timeout 10.0))
(,prog (and (stringp ,prog-msg)
(make-progress-reporter ,prog-msg))))
(with-timeout (,(pop spec) ,@spec)
(with-timeout (,timeout ,@spec)
(while (not (setq ,res (progn ,@wait-forms)))
(accept-process-output nil 0.0001)
(accept-process-output nil ,wait-time)
(when ,prog (progress-reporter-update ,prog))))
(prog1 ,res
(when ,prog (progress-reporter-done ,prog))))))