"is" And "id" In Python 3.5
Solution 1:
id
returns a Python int
object (the memory address of the object you're id
-ing, though that's an implementation detail). But aside from very small int
s (again, implementation detail), there is no int
caching in Python; if you compute the same int
two ways, it's two different int
objects that happen to have the same value. Similarly, a fresh int
is created on each call to id
, even if the objects are the same.
The equivalence for id
and is
is that a is b
implies id(a) == id(b)
, not id(a) is id(b)
(and in fact, since id
s are large numbers, id(a) is id(b)
is almost always False
).
Also note, your test case is flawed in other ways:
a = 3
b = 3
a is b
only returns True
for the is
comparison because of the small int
cache in CPython; if you'd done:
a = 1000
b = 1000
a is b
a is b
would be False
; your assumptions about identity only hold in CPython for numbers in the range -5 to 256 inclusive, which are singletons for performance reasons, but all other int
s are recreated as needed, not singletons.
Solution 2:
id(a) is id(b)
compares the ids of the ids returned by the id
function.
id(a) == id(b)
will be True
since the a is b
and during an object's lifetime, the value returned by id
will always be the same. However, each time id
is called, a distinct integer (that has the same value) is returned so id(a) is id(b)
is False
.
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