mirror of
https://github.com/python/cpython.git
synced 2024-12-25 01:34:48 +08:00
152 lines
3.7 KiB
Python
152 lines
3.7 KiB
Python
doctests = """
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########### Tests mostly copied from test_listcomps.py ############
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Test simple loop with conditional
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>>> sum({i*i for i in range(100) if i&1 == 1})
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166650
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Test simple case
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>>> {2*y + x + 1 for x in (0,) for y in (1,)}
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{3}
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Test simple nesting
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>>> list(sorted({(i,j) for i in range(3) for j in range(4)}))
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[(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3)]
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Test nesting with the inner expression dependent on the outer
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>>> list(sorted({(i,j) for i in range(4) for j in range(i)}))
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[(1, 0), (2, 0), (2, 1), (3, 0), (3, 1), (3, 2)]
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Make sure the induction variable is not exposed
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>>> i = 20
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>>> sum({i*i for i in range(100)})
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328350
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>>> i
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20
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Verify that syntax error's are raised for setcomps used as lvalues
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>>> {y for y in (1,2)} = 10 # doctest: +IGNORE_EXCEPTION_DETAIL
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Traceback (most recent call last):
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...
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SyntaxError: ...
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>>> {y for y in (1,2)} += 10 # doctest: +IGNORE_EXCEPTION_DETAIL
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Traceback (most recent call last):
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...
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SyntaxError: ...
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Make a nested set comprehension that acts like set(range())
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>>> def srange(n):
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... return {i for i in range(n)}
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>>> list(sorted(srange(10)))
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[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
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Same again, only as a lambda expression instead of a function definition
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>>> lrange = lambda n: {i for i in range(n)}
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>>> list(sorted(lrange(10)))
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[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
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Generators can call other generators:
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>>> def grange(n):
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... for x in {i for i in range(n)}:
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... yield x
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>>> list(sorted(grange(5)))
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[0, 1, 2, 3, 4]
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Make sure that None is a valid return value
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>>> {None for i in range(10)}
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{None}
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########### Tests for various scoping corner cases ############
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Return lambdas that use the iteration variable as a default argument
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>>> items = {(lambda i=i: i) for i in range(5)}
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>>> {x() for x in items} == set(range(5))
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True
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Same again, only this time as a closure variable
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>>> items = {(lambda: i) for i in range(5)}
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>>> {x() for x in items}
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{4}
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Another way to test that the iteration variable is local to the list comp
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>>> items = {(lambda: i) for i in range(5)}
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>>> i = 20
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>>> {x() for x in items}
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{4}
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And confirm that a closure can jump over the list comp scope
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>>> items = {(lambda: y) for i in range(5)}
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>>> y = 2
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>>> {x() for x in items}
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{2}
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We also repeat each of the above scoping tests inside a function
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>>> def test_func():
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... items = {(lambda i=i: i) for i in range(5)}
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... return {x() for x in items}
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>>> test_func() == set(range(5))
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True
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>>> def test_func():
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... items = {(lambda: i) for i in range(5)}
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... return {x() for x in items}
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>>> test_func()
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{4}
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>>> def test_func():
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... items = {(lambda: i) for i in range(5)}
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... i = 20
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... return {x() for x in items}
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>>> test_func()
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{4}
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>>> def test_func():
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... items = {(lambda: y) for i in range(5)}
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... y = 2
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... return {x() for x in items}
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>>> test_func()
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{2}
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"""
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__test__ = {'doctests' : doctests}
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def test_main(verbose=None):
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import sys
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from test import support
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from test import test_setcomps
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support.run_doctest(test_setcomps, verbose)
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# verify reference counting
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if verbose and hasattr(sys, "gettotalrefcount"):
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import gc
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counts = [None] * 5
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for i in range(len(counts)):
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support.run_doctest(test_setcomps, verbose)
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gc.collect()
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counts[i] = sys.gettotalrefcount()
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print(counts)
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if __name__ == "__main__":
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test_main(verbose=True)
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