On May 31, 12:31 am, "Warren Stringer" <[EMAIL PROTECTED]> wrote: > This is inconsistent: > > why does c[:][0]() work but c[:]() does not? > Why does c[0]() has exactly the same results as c[:][0]() ? > Moreover, c[:][0]() implies that a slice was invoked
It's not inconsistent, but [:] probably does something different than you think it does. All it does is create a copy (not in general, but at least if c is a list or a tuple). Since in your example c is a tuple and tuples are immutable, making a copy of it is essentially useless. Why not just use the original? I.e. instead of c[:] you could just write c. That's why c[:][0]() has exactly the same effect as c[0] (), although the former is likely to be slightly slower. c[:]() tries to call the copied tuple. Tuples aren't callable. c[:][0]() calls the first element in the copied tuple, and that element happens to be callable. -- http://mail.python.org/mailman/listinfo/python-list
