Question

Why does

sys.getrefcount() 

return 3 for every large number or simple string?Does that mean that 3 objects reside somewhere in the Program?Also,why doesn’t setting x=(very large number) increase that object’s ref count?Do those 3 ref counts result from my call to getrefcount?
Thank you for clarifying this.

for instance:

>>> sys.getrefcount(4234234555)
3
>>> sys.getrefcount("testing")
3
>>> sys.getrefcount(11111111111111111)
3
>>> x=11111111111111111
>>> sys.getrefcount(11111111111111111)
3 

Answer 1

Large integer objects are not reused by the interpretor, so you get two distinct objects:

>>> a = 11111
>>> b = 11111
>>> id(a)
40351656
>>> id(b)
40351704

sys.getrefcount(11111) always returns the same number because it measures the reference count of a fresh object.

For small integers, Python always reuses the same object:

>>> sys.getrefcount(1)
73

Usually you would get only one reference to a new object:

>>> sys.getrefcount(object())
1

But integers are allocated in a special pre-malloced area by Python for performance optimization, and I suspect the extra two references have something to do with this.

It’s implemented in longobject.c in CPython. (Update: link to Python3.) I do not claim to understand what’s really going on. I think there are several things at work that cache temporary references:

print sys.getrefcount('foo1111111111111' + 'bar1111111111111') #1
print sys.getrefcount(111111111111 + 2222222222222)            #2
print sys.getrefcount('foobar333333333333333333')              #3