CODEWITHSUNDEEP
pythonstringidentitypython-internals
Chris_Rands
Chris_Rands

Reputation: 41198

sys.intern() failing to intern a string upon slicing

Why is sys.intern() failing to intern this string?

>>> from sys import intern
>>> "abcd"[:-1] == "abc"
True
>>> "abcd"[:-1] is "abc"
False
>>> intern("abc")
'abc'
>>> "abcd"[:-1] is "abc"
False  # Expected True

(on CPython 3.7.4)

Upvotes: 2

Views: 110

Answers (2)

ead
ead

Reputation: 34367

Your expectations are wrong. Names, string-constants and similar are interned when the code-object is created (here) and not when any string-object is created.

The string for "abcd"[:-1] is created during the run time and thus not interned. Theoretically, if optimizer would evaluate this expression during the compile time, it could be interned, as it happens for example with

>>> "a"*5 is "aaaaa"  #True

But it is just an impelementation detail, for example

>>> "a"*4097

would no longer be evaluated during the compile time and thus not interned.

Upvotes: 0

Alex Hall
Alex Hall

Reputation: 36043

String literals are already interned, it's computed expressions that you'd need to intern manually.

from sys import intern
print("abcd"[:-1] is "abc")  # False
print(intern("abcd"[:-1]) is "abc")  # True
print("abcd"[:-1] is "abc")  # False

intern doesn't mean 'whenever this string gets produced anywhere in the program change it to the interned reference', it just returns the interned reference for the string it's given.

s1 = "abc"
s2 = "abc"
s3 = "abcd"[:-1]
s4 = intern(s3)
for s in [s1, s2, s3, s4]:
  print(id(s))

140002991598512
140002991598512
140002990838576
140002991598512

Upvotes: 3

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