New submission from Albert <albert....@gmail.com>: Hello!
Is it intentional, that the default argument is ALWAYS evaluated, even if it is not needed??? Is it not a bug, that this method has no short-circuit eval (http://en.wikipedia.org/wiki/Short-circuit_evaluation) ?? Example1: ========= infinite = 1e100 one_div_by = {0.0 : infinite} def func(n): return one_div_by.setdefault(float(n), 1/float(n)) for i in [1, 2, 3, 4]: print i, func(i) print one_div_by # works!! for i in [0, 1, 2, 3, 4]: # added 0 -> FAIL! print i, func(i) print one_div_by # fail!! Example2: ========= fib_d = {0 : 0, 1 : 1} def fibonacci(n): return fib_d.setdefault(n, fibonacci(n-1) + fibonacci(n-2)) for i in range(10): print i, fibonacci(i) print fib_d ---------- messages: 126456 nosy: albert.neu priority: normal severity: normal status: open title: dict.setdefault: Bug: default argument is ALWAYS evaluated, i.e. no short-circuit eval type: behavior versions: Python 2.6 _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue10930> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
