Matt Hammond wrote: > On Fri, 12 Aug 2005 12:57:38 +0100, Peter Mott <[EMAIL PROTECTED]> wrote: > >> If I use concatenation + instead of multiplication * then I get the >> result that Jiri expected: >> >> >>> L = [[]] + [[]] >> >>> L[1].append(1) >> >>> L >> [[], [1]] >> >> With * both elements are changed: >> >> >>> L = [[]] * 2 >> >>> L[1].append(1) >> >>> L >> [[1], [1]] >> >> Alex Martelli says in his excellent Nutshell book that + is >> concatenation and that "n*S is the concatenation of n copies of S". >> But it seems not so. Surely, from a logical point of view, S + S >> should be the same as S * 2? > > > S+S is the same as S*2, but L= [[]] + [[]] is not S+S. The two terms > being added are different instances of an empty list. You are > adding/concatenating two different object instances.
But it is still true that [[]] + [[]] is not the same as [[]] * 2. In my usage anyway this means that "S+S is the same as S*2" is false. Because there are Python expressions for which it is falsfied. The problem I have is pretty philosophical I admit, but I don't think you do it justice. It's really about identity. Logically speaking identity is a congruence relation of a language, which means that if x=y is true then C(x) = C(y) will be true for any context C of the lanuage. Python is so regular that most of the time it follows this. But not with *. If you define C(x): def C(x): ... x[1].append(1) ... return x[0] then although [[]]+[[]] == [[]]*2 evaluates true C([[]]+[[]]) is different from C([[]]*2). So == is not a congruence in Python. Inbteresting that Phil Hunt just posted about 'Unify' which, if I understand it right, has the feature that provided expressions S and T are in the canonical "Storage Manager" format then == will be a congruence and hence an idenity. Cheers Peter -- http://mail.python.org/mailman/listinfo/python-list
