47 lines
1.0 KiB
Python
47 lines
1.0 KiB
Python
# stress test for the heap by allocating lots of objects within threads
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# allocates about 5mb on the heap
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#
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# MIT license; Copyright (c) 2016 Damien P. George on behalf of Pycom Ltd
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import time
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import _thread
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def last(l):
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return l[-1]
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def thread_entry(n):
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# allocate a bytearray and fill it
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data = bytearray(i for i in range(256))
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# run a loop which allocates a small list and uses it each iteration
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lst = 8 * [0]
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sum = 0
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for i in range(n):
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sum += last(lst)
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lst = [0, 0, 0, 0, 0, 0, 0, i + 1]
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# check that the bytearray still has the right data
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for i, b in enumerate(data):
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assert i == b
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# print the result of the loop and indicate we are finished
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with lock:
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print(sum, lst[-1])
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global n_finished
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n_finished += 1
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lock = _thread.allocate_lock()
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n_thread = 10
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n_finished = 0
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# spawn threads
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for i in range(n_thread):
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_thread.start_new_thread(thread_entry, (10000,))
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# wait for threads to finish
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while n_finished < n_thread:
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time.sleep(1)
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