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When is a python object's hash computed and why is the hash of -1 different?

Why do -1 and -2 both hash to the same number if Python?

Since they do, how does Python tell these two numbers apart?

>>> -1 is -2
False
>>> hash(-1) is hash(-2)
True
>>> hash(-1)
-2
>>> hash(-2)
-2

Answer 1

-1 is a reserved value at the C level of CPython which prevents hash functions from being able to produce a hash value of -1. As noted by DSM, the same is not true in IronPython and PyPy where hash(-1) != hash(-2).

See this Quora answer:

If you write a type in a C extension module and provide a tp_hash
method, you have to avoid -1 — if you return -1, Python will assume
you meant to throw an error.

If you write a class in pure Python and provide a __hash__ method,
there’s no such requirement, thankfully. But that’s because the C code
that invokes your __hash__ method does that for you — if your
__hash__ returns -1, then hash() applied to your object will actually return -2.

Which really just repackages the information from effbot:

The hash value -1 is reserved (it’s used to flag errors in the C
implementation). If the hash algorithm generates this value, we simply
use -2 instead.

You can also see this in the source. For example for Python 3’s int object, this is at the end of the hash implementation:

if (x == (Py_uhash_t)-1)
    x = (Py_uhash_t)-2;
return (Py_hash_t)x;

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Since they do, how does Python tell these two numbers apart?

Since all hash functions map a large input space to a smaller input space, collisions are always expected, no matter how good the hash function is. Think of hashing strings, for example. If hash codes are 32-bit integers, you have 2^32 (a little more than 4 billion) hash codes. If you consider all ASCII strings of length 6, you have (2^7)^6 (just under 4.4 trillion) different items in your input space. With only this set, you are guaranteed to have many, many collisions no matter how good you are. Add Unicode characters and strings of unlimited length to that!

Therefore, the hash code only hints at the location of an object, an equality test follows to test candidate keys. To implement a membership test in a hash-table set, the hash code gives you “bucket” number in which to search for the value. However, all set items with the same hash code are in the bucket. For this, you also need an equality test to distinguish between all candidates in the bucket.

This hash code and equality duality is hinted at in the CPython documentation on hashable objects. In other languages/frameworks, there is a guideline/rule that if you provide a custom hash code function, you must also provide a custom equality test (performed on the same fields as the hash code function).

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Indeed, the Python release today address exactly this, with a security patch that addresses the efficiency issue when this (identical hash values, but on a massive scale) is used as a denial of service attack – http://mail.python.org/pipermail/python-list/2012-April/1290792.html