Terry J. Reedy <tjre...@udel.edu> added the comment:
Another possibility might be to not use the Windows timeout clock, at least not
for short timeouts. The following shows that tk does about 970 1 millesecond
timeouts in 1 second (on my machine).
---
import tkinter as tk
root = tk.Tk()
cbid = None
cbcount = 0
def cb():
global cbid, cbcount
cbcount += 1
cbid = root.after(1, cb)
def cbcancel():
print(cbcount)
root.after_cancel(cbid)
root.after(1000, cbcancel)
cbid = root.after(1, cb)
root.mainloop()
---
Here is a proof-of-concept queue-get-with-timeout function with sub-millesecond
resolution.
---
import tkinter as tk
from queue import Queue, Empty
from time import perf_counter
q = Queue()
fails = 0
value = '<raise Empty>'
def qcb():
try:
global value
value = q.get(block=False)
except Empty:
global fails
fails += 1
if perf_counter() < stop:
root.after(0, qcb)
return
root.destroy()
def qget_to(timeout):
global root, stop
root = tk.Tk()
root.withdraw()
stop = perf_counter() + timeout
qid = root.after(0, qcb)
#q.put(1)
root.mainloop()
print('failures:', fails, 'value:', value)
qget_to(.001)
---
With the put commented out, there are 27 fails (on my machine). When a value
is already available, there are none. The loop could be parameterized to
repeatedly call the no-block version of any similar function. Large enough
time-outs could be partially fulfilled using the system timeout function.
----------
nosy: +terry.reedy
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue34535>
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