Why does `None is None is None` return True? [duplicate]

Answer 1

The bytecode shows that two comparisons are being performed here with the middle being duplicated:

>>> import dis
>>> def a():
...     return None is None is None
... 
>>> dis.dis(a)
  2           0 LOAD_CONST               0 (None)
              3 LOAD_CONST               0 (None)
              6 DUP_TOP
              7 ROT_THREE
              8 COMPARE_OP               8 (is)
             11 JUMP_IF_FALSE_OR_POP    21
             14 LOAD_CONST               0 (None)
             17 COMPARE_OP               8 (is)
             20 RETURN_VALUE
        >>   21 ROT_TWO
             22 POP_TOP
             23 RETURN_VALUE

As stated in the docs for comparisons this is because these operators chain together.

a op b op c will be translated to a op b and b op c (note b is duplicated in the bytecode as shown above)

Answer 2

As some people comments, Python comparisons can be chained.

For the sake of explanation, when chaining, Python actually ANDs the expressions.

The rationale behind this, is that expressions like a < b < c have the interpretation that is conventional in mathematics. Hence the confusion of your particular expression None is None is None where identy operators are involved.

So basically, this would translate to:

(None is None) and (None is None)

which is clearly True

Here is another example in the Python docs

Further Information

Especially since this was an interview question, it is important to note that this is not a general behavior shared among all languages.

As it is stated in the documentation I linked,

Unlike C, all comparison operations in Python have the same priority, which is lower than that of any arithmetic, shifting or bitwise operation.

So, let's consider the 10 > x > 2 expression (since is operator is not valid in C).

C's translation (because of operator precedence)

((10 > x) > 2)

Python's translation

(10 > x) and (x > 2)

Answer 3

is is a comparison operator, as seen in the docs:

comparison    ::=  or_expr ( comp_operator or_expr )*
comp_operator ::=  "<" | ">" | "==" | ">=" | "<=" | "!="
                   | "is" ["not"] | ["not"] "in"

So just like the other comparison operators, it can be chained arbitrarily. So

a = b = c = None
a is b is c

is equivalent to

(a is b) and (b is c)