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.

>>> "string" is "string" True 

This is as I expected.

You can also do this.

>>> "strin"+"g" is "string" True 

And that's pretty clever!

But you can't do this.

>>> s1 = "strin" >>> s2 = "string" >>> s1+"g" is s2 False 

Why wouldn't Python evaluate s1+"g", and realize it is the same as s2 and point it to the same address? What is actually going on in that last block to have it return False?

Answer 1

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

In what follows CPython 3.9.0+ is used.

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"   1           0 LOAD_CONST               0 ('string')               2 STORE_NAME               0 (s1)    # s2 = "strin" + "g"   2           4 LOAD_CONST               0 ('string')               6 STORE_NAME               1 (s2) 

This bytecode was obtained with (which prints a few more lines after the above):

import dis  source = 's1 = "string"\ns2 = "strin" + "g"' code = compile(source, '', 'exec') print(dis.dis(code)) 

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

  # s3a = "strin"   3           8 LOAD_CONST               1 ('strin')              10 STORE_NAME               2 (s3a)    # s3 = s3a + "g"   4          12 LOAD_NAME                2 (s3a)              14 LOAD_CONST               2 ('g')              16 BINARY_ADD              18 STORE_NAME               3 (s3)              20 LOAD_CONST               3 (None)              22 RETURN_VALUE 

This bytecode was obtained with (which prints a few more lines before the above, and those lines are exactly as in the first block of bytecodes given above):

import dis  source = (     's1 = "string"\n'     's2 = "strin" + "g"\n'     's3a = "strin"\n'     's3 = s3a + "g"') code = compile(source, '', 'exec') print(dis.dis(code)) 

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

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

Also, Python 3.9 prints a warning for the last two statements above:

SyntaxWarning: "is" with a literal. Did you mean "=="?

Answer 2

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