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Sun May 23 20:40:21 EDT 2004 · https://mail.python.org/pipermail/python-dev/2004-May/044942.html

Gustavo Niemeyer <niemeyer at conectiva.com>:

> Is it true that the current list comparison algorithm
> goes trough every element doing identity comparison, and
> if every element in the first list *is* the element at
> the same position in the second list, and the list has
> the same size, then lists are considered equal?

> If that's true, then I belive there's no reason for
> not comparing if the first list *is* the second list at
> the top of list_richcompare(), right?

What if the list contains a NaN?

Greg Ewing, Computer Science Dept, +--------------------------------------+
University of Canterbury,	   | A citizen of NewZealandCorp, a	  |
Christchurch, New Zealand	   | wholly-owned subsidiary of USA Inc.  |
greg at cosc.canterbury.ac.nz	   +--------------------------------------+

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