[Python-Dev] PyObject_RichCompareBool identity shortcut (original) (raw)

Mark Dickinson dickinsm at gmail.com
Wed Apr 27 23:15:46 CEST 2011

On Wed, Apr 27, 2011 at 7:41 PM, Glenn Linderman <v+python at g.nevcal.com> wrote:

One issue that I don't fully understand: I know there is only one instance of None in Python, but I'm not sure where to discover whether there is only a single, or whether there can be multiple, instances of NaN or Inf.  The IEEE 754 spec is clear that there are multiple bit sequences that can be used to represent these, so I would hope that there can be, in fact, more than one value containing NaN (and Inf).

This would properly imply that a collection should correctly handle the case of storing multiple, different items using different NaN (and Inf) instances.  A dict, for example, should be able to hold hundreds of items with the index value of NaN. The distinction between "is" and "==" would permit proper operation, and I believe that Python's "rebinding" of names to values rather than the copying of values to variables makes such a distinction possible to use in a correct manner.

For infinities, there's no issue: there are exactly two distinct infinities (+inf and -inf), and they don't have any special properties that affect membership tests. Your float-keyed dict can contain both +inf and -inf keys, or just one, or neither, in exactly the same way that it can contain both +5.0 and -5.0 as keys, or just one, or neither.

For nans, you can put multiple nans into a dictionary as separate keys, but under the current rules the test for 'sameness' of two nan keys becomes a test of object identity, not of bitwise equality. Python takes no notice of the sign bits and 'payload' bits of a float nan, except in operations like struct.pack and struct.unpack. For example:

x, y = float('nan'), float('nan') d = {x: 1, y:2} x in d True y in d True d[x] 1 d[y] 2

But using struct.pack, you can see that x and y are bitwise identical:

struct.pack('<d', x) == struct.pack('<d', y) True

Mark

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