Python string interning

Question

While this question doesn\'t have any real use in practice, I am curious as to how Python does string interning. I have noticed the following.

>>> \         


        

Answer 1

Case 1

>>> x = "123"  
>>> y = "123"  
>>> x == y  
True  
>>> x is y  
True  
>>> id(x)  
50986112  
>>> id(y)  
50986112  

Case 2

>>> x = "12"
>>> y = "123"
>>> x = x + "3"
>>> x is y
False
>>> x == y
True

Now, your question is why the id is same in case 1 and not in case 2.
In case 1, you have assigned a string literal "123" to x and y.

Since string are immutable, it makes sense for the interpreter to store the string literal only once and point all the variables to the same object.
Hence you see the id as identical.

In case 2, you are modifying x using concatenation. Both x and y has same values, but not same identity.
Both points to different objects in memory. Hence they have different id and is operator returned False

Answer 2

This is implementation-specific, but your interpreter is probably interning compile-time constants but not the results of run-time expressions.

In what follows I use CPython 2.7.3.

In the second example, the expression "strin"+"g" is evaluated at compile time, and is replaced with "string". This makes the first two examples behave the same.

If we examine the bytecodes, we'll see that they are exactly the same:

  # s1 = "string"
  2           0 LOAD_CONST               1 ('string')
              3 STORE_FAST               0 (s1)

  # s2 = "strin" + "g"
  3           6 LOAD_CONST               4 ('string')
              9 STORE_FAST               1 (s2)

The third example involves a run-time concatenation, the result of which is not automatically interned:

  # s3a = "strin"
  # s3 = s3a + "g"
  4          12 LOAD_CONST               2 ('strin')
             15 STORE_FAST               2 (s3a)

  5          18 LOAD_FAST                2 (s3a)
             21 LOAD_CONST               3 ('g')
             24 BINARY_ADD          
             25 STORE_FAST               3 (s3)
             28 LOAD_CONST               0 (None)
             31 RETURN_VALUE        

If you were to manually intern() the result of the third expression, you'd get the same object as before:

>>> s3a = "strin"
>>> s3 = s3a + "g"
>>> s3 is "string"
False
>>> intern(s3) is "string"
True