Douglas Drumond wrote: > > In a2() you do l1 += l2, ie, l1 = l1 + l2
A subtle clarification is warranted I think. l1 += l2 is not the same as l1 = l1 + l2, when l1 and l2 are lists. l1 += l2 is an augmented assignment statement, and as such will perform the operation in place if possible, IIUC. Consider the following: In [1]: l1 = l2 = [1, 2, 3] In [2]: l1 is l2 Out[2]: True In [3]: l1 += [4, 5, 6] # in-place In [4]: l1 is l2 Out[4]: True In [5]: l1 = l1 + [7, 8, 9] # not In [6]: l1 is l2 Out[6]: False Perhaps there is a better reference, but this behavior is discussed briefly here: http://docs.python.org/ref/augassign.html > But if you don't have l1 defined yet, you can't add to l2 > It's like: > def a2(): > l1 = foo + l2 > > > UnboundLocalError: local variable 'foo' referenced before assignment > > It's because l1 (and foo at above example) is a local variable. > a1's l1 is different from a2's l1. Yes, but as Alan pointed out it's considered local because of the assignment attempt. Obligatory doc reference: http://www.python.org/doc/2.4/ref/naming.html snip = """\ If a name binding operation occurs anywhere within a code block, all uses of the name within the block are treated as references to the current block. This can lead to errors when a name is used within a block before it is bound. This rule is subtle. Python lacks declarations and allows name binding operations to occur anywhere within a code block. The local variables of a code block can be determined by scanning the entire text of the block for name binding operations. """ HTH, Marty _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
