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docs/utime: Remove only:: for ticks_diff(). 

It's mandatory function which should be present in every port. Even if
it's not, in the stdlib intro we waarn users that a particular port can
lack anything of described in the docs.
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Paul Sokolovsky 2016-11-01 00:03:40 +03:00
parent 7ffc959c00
commit 8679d9e6a6
1 changed files with 43 additions and 45 deletions
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@ -116,59 +116,57 @@ Functions
Availability: Not every port implement this function.
.. only:: port_unix or port_pyboard or port_wipy or port_esp8266
.. function:: ticks_diff(ticks1, ticks2)
.. function:: ticks_diff(ticks1, ticks2)
Measure ticks difference between values returned from ticks_ms(), ticks_us(), or ticks_cpu()
functions. The argument order is the same as for subtraction operator,
``tick_diff(ticks1, ticks2)`` has the same meaning as ``ticks1 - ticks2``. However, values returned by
ticks_ms(), etc. functions may wrap around, so directly using subtraction on them will
produce incorrect result. That is why ticks_diff() is needed, it implements modular
(or more specifically, ring) arithmetics to produce correct result even for wrap-around
values (as long as they not too distant inbetween, see below). The function returns
**signed** value in the range [`-TICKS_PERIOD/2` .. `TICKS_PERIOD/2-1`] (that's a typical
range definition for two's-complement signed binary integers). If the result is negative,
it means that `ticks1` occured earlier in time than `ticks2`. Otherwise, it means that
`ticks1` was after `ticks2`. This holds `only` if `ticks1` and `ticks2` are apart from
each other for no more than `TICKS_PERIOD/2-1` ticks. If that does not hold, incorrect
result will be returned. Specifically, if 2 tick values are apart for `TICKS_PERIOD/2-1`
ticks, that value will be returned by the function. However, if `TICKS_PERIOD/2` of
real-time ticks has passed between them, the function will return `-TICKS_PERIOD/2`
instead, i.e. result value will wrap around to the negative range of possible values.
Measure ticks difference between values returned from ticks_ms(), ticks_us(), or ticks_cpu()
functions. The argument order is the same as for subtraction operator,
``tick_diff(ticks1, ticks2)`` has the same meaning as ``ticks1 - ticks2``. However, values returned by
ticks_ms(), etc. functions may wrap around, so directly using subtraction on them will
produce incorrect result. That is why ticks_diff() is needed, it implements modular
(or more specifically, ring) arithmetics to produce correct result even for wrap-around
values (as long as they not too distant inbetween, see below). The function returns
**signed** value in the range [`-TICKS_PERIOD/2` .. `TICKS_PERIOD/2-1`] (that's a typical
range definition for two's-complement signed binary integers). If the result is negative,
it means that `ticks1` occured earlier in time than `ticks2`. Otherwise, it means that
`ticks1` was after `ticks2`. This holds `only` if `ticks1` and `ticks2` are apart from
each other for no more than `TICKS_PERIOD/2-1` ticks. If that does not hold, incorrect
result will be returned. Specifically, if 2 tick values are apart for `TICKS_PERIOD/2-1`
ticks, that value will be returned by the function. However, if `TICKS_PERIOD/2` of
real-time ticks has passed between them, the function will return `-TICKS_PERIOD/2`
instead, i.e. result value will wrap around to the negative range of possible values.
``ticks_diff()`` is designed to accommodate various usage patterns, among them:
``ticks_diff()`` is designed to accommodate various usage patterns, among them:
Polling with timeout. In this case, the order of events is known, and you will deal
only with positive results of ``ticks_diff()``::
Polling with timeout. In this case, the order of events is known, and you will deal
only with positive results of ``ticks_diff()``::
# Wait for GPIO pin to be asserted, but at most 500us
start = time.ticks_us()
while pin.value() == 0:
if time.ticks_diff(time.ticks_us(), start) > 500:
raise TimeoutError
# Wait for GPIO pin to be asserted, but at most 500us
start = time.ticks_us()
while pin.value() == 0:
if time.ticks_diff(time.ticks_us(), start) > 500:
raise TimeoutError
Scheduling events. In this case, ``ticks_diff()`` result may be negative
if an event is overdue::
Scheduling events. In this case, ``ticks_diff()`` result may be negative
if an event is overdue::
# This code snippet is not optimized
now = time.ticks_ms()
scheduled_time = task.scheduled_time()
if ticks_diff(now, scheduled_time) > 0:
print("Too early, let's nap")
sleep_ms(ticks_diff(now, scheduled_time))
task.run()
elif ticks_diff(now, scheduled_time) == 0:
print("Right at time!")
task.run()
elif ticks_diff(now, scheduled_time) < 0:
print("Oops, running late, tell task to run faster!")
task.run(run_faster=true)
# This code snippet is not optimized
now = time.ticks_ms()
scheduled_time = task.scheduled_time()
if ticks_diff(now, scheduled_time) > 0:
print("Too early, let's nap")
sleep_ms(ticks_diff(now, scheduled_time))
task.run()
elif ticks_diff(now, scheduled_time) == 0:
print("Right at time!")
task.run()
elif ticks_diff(now, scheduled_time) < 0:
print("Oops, running late, tell task to run faster!")
task.run(run_faster=true)
Note: Do not pass ``time()`` values to ``ticks_diff()``, and should use
normal mathematical operations on them. But note that ``time()`` may (and will)
also overflow. This is known as https://en.wikipedia.org/wiki/Year_2038_problem .
Note: Do not pass ``time()`` values to ``ticks_diff()``, and should use
normal mathematical operations on them. But note that ``time()`` may (and will)
also overflow. This is known as https://en.wikipedia.org/wiki/Year_2038_problem .
.. function:: time()