2014-08-29 15:05:32 -04:00
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# test if conditions which are optimised by the compiler
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2014-10-17 13:57:33 -04:00
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if 0:
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print(5)
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else:
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print(6)
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if 1:
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print(7)
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if 2:
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print(8)
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if -1:
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print(9)
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elif 1:
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print(10)
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if 0:
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print(11)
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else:
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print(12)
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if 0:
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print(13)
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elif 1:
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print(14)
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if 0:
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print(15)
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elif 0:
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print(16)
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else:
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print(17)
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2015-03-12 18:47:44 -04:00
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if not False:
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print('a')
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if not True:
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print('a')
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else:
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print('b')
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if False:
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print('a')
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else:
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print('b')
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if True:
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print('a')
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if (1,):
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print('a')
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if not (1,):
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print('a')
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else:
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print('b')
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py/compile: Remove over-eager optimisation of tuples as if condition.
When a tuple is the condition of an if statement, it's only possible to
optimise that tuple away when it is a constant tuple (ie all its elements
are constants), because if it's not constant then the elements must be
evaluated in case they have side effects (even though the resulting tuple
will always be "true").
The code before this change handled the empty tuple OK (because it doesn't
need to be evaluated), but it discarded non-empty tuples without evaluating
them, which is incorrect behaviour (as show by the updated test).
This optimisation is anyway rarely applied because it's not common Python
coding practice to write things like `if (): ...` and `if (1, 2): ...`, so
removing this optimisation completely won't affect much code, if any.
Furthermore, when MICROPY_COMP_CONST_TUPLE is enabled, constant tuples are
already optimised by the parser, so expression with constant tuples like
`if (): ...` and `if (1, 2): ...` will continue to be optimised properly
(and so when this option is enabled the code that's deleted in this commit
is actually unreachable when the if condition is a constant tuple).
Signed-off-by: Damien George <damien@micropython.org>
2023-05-02 23:21:18 -04:00
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# test evaluation of the if-condition with tuples as arguments
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# non-constant tuples should be evaluated even though they will evaluate to true
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def f(x):
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print("f", x)
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if (f(1),):
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print(18)
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if (f(2), f(3)):
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print(19)
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# test if-conditions within a function
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2014-08-29 15:05:32 -04:00
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f2 = 0
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def f(t1, t2, f1):
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if False:
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print(1)
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if True:
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print(1)
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if ():
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print(1)
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if (1,):
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print(1)
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if (1, 2):
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print(1)
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if t1 and t2:
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print(1)
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if (t1 and t2): # parsed differently to above
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print(1)
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if not (t1 and f1):
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print(1)
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if t1 or t2:
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print(1)
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if (t1 or t2): # parse differently to above
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print(1)
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if f1 or t1:
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print(1)
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if not (f1 or f2):
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print(1)
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if t1 and f1 or t1 and t2:
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print(1)
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if (f1 or t1) and (f2 or t2):
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print(1)
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f(True, 1, False)
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