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adminpete 2015-11-29 17:37:41 +00:00
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LICENSE 100644
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The MIT License (MIT)
Copyright (c) 2014 Peter Hinch
Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
SOFTWARE.

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mutex/README.md 100644
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# mutex.py mutex_test.py
Implementation of a very simple mutex class
[reference](http://www.geeksforgeeks.org/mutex-vs-semaphore/)

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mutex/mutex.py 100644
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import pyb, micropython, array, uctypes
micropython.alloc_emergency_exception_buf(100)
class MutexException(OSError):
pass
class Mutex:
@micropython.asm_thumb
def _acquire(r0, r1): # Spinlock: wait on the semaphore. Return on success.
label(LOOP)
ldr(r0, [r1, 0]) # Wait for lock to be zero
cmp(r0, 0)
bne(LOOP) # Another process has the lock: spin on it
cpsid(0) # OK, we have lock at this instant disable interrupts
ldr(r0, [r1, 0]) # and re-check in case an interrupt occurred
cmp(r0, 0)
itt(ne) # if someone got in first re-enable ints
cpsie(0) # and start polling again
b(LOOP)
mov(r0, 1) # We have an exclusive access
str(r0, [r1, 0]) # set the lock
cpsie(0)
@micropython.asm_thumb
def _attempt(r0, r1): # Nonblocking. Try to lock. Return 0 on success, 1 on fail
cpsid(0) # disable interrupts
ldr(r0, [r1, 0])
cmp(r0, 0)
bne(FAIL) # Another process has the lock: fail
mov(r2, 1) # No lock
str(r2, [r1, 0]) # set the lock
label(FAIL)
cpsie(0) # enable interrupts
def __init__(self):
self.lock = array.array('i', (0,)) # 1 if a process has the lock else 0
# POSIX API pthread_mutex_lock() blocks the thread till resource is available.
def __enter__(self):
self._acquire(uctypes.addressof(self.lock))
return self
def __exit__(self, *_):
self.lock[0] = 0
# POSIX pthread_mutex_unlock()
def release(self):
if self.lock[0] == 0:
raise MutexException('Semaphore already released')
self.lock[0] = 0
# POSIX pthread_mutex_trylock() API. When mutex is not available the function returns immediately
def test(self): # Nonblocking: try to acquire, return True if success.
return self._attempt(uctypes.addressof(self.lock)) == 0

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import pyb, mutex
mutex = mutex.Mutex()
def update(t): # Interrupt handler may not be able to acquire the lock
global var1, var2 # if main loop has it
if mutex.test(): # critical section start
var1 += 1
pyb.udelay(200)
var2 += 1
mutex.release() # critical section end
def main():
global var1, var2
var1, var2 = 0, 0
t2 = pyb.Timer(2, freq = 995, callback = update)
t4 = pyb.Timer(4, freq = 1000, callback = update)
for x in range(1000000):
with mutex: # critical section start
a = var1
pyb.udelay(200)
b = var2
result = a == b # critical section end
if not result:
print('Fail after {} iterations'.format(x))
break
pyb.delay(1)
if x % 1000 == 0:
print(x)
t2.deinit()
t4.deinit()
print(var1, var2)