Re: Modifying the default argument of function

On Tue, 21 Jan 2014 20:11:02 +0100 Mû m...@melix.net wrote: Hi everybody, A friend of mine asked me a question about the following code: [code] def f(x=[2,3]): x.append(1) return x print(f()) print(f()) print(f()) [/code] The results are [2, 3, 1], [2, 3, 1, 1] and

[issue1500] Example using range doesn't give claimed results

New submission from Steve Jones: See for yourself, on http://docs.python.org/tut/node6.html * for n in range(2, 10): ... for x in range(2, n): ... if n % x == 0: ... print n, 'equals', x, '*', n/x ... break ... else