Python string interning -

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While there is no real use in the behavior of this question, I'm curious how the Python string interns. I have noticed the following. 

  & gt; & Gt; "String" is "string" & gt; & Gt; This is true   
 It was just like I expected.  
 You can also do this.  
  & gt; & Gt; "String" + "G" is "string" & gt; & Gt;  
  
 S1 = "strin" & gt; & Gt; S2 = "string" & gt; & Gt; S1 + "g" is s2 & gt; & Gt; Incorrect   
 Python does not have an evaluation of  s1 + "g" , it is realized that it is like  s1  and it is the same Point to the address? In fact, what is happening in that last section from which it is  incorrect    
 
 
 Implementation is specific, but your interpreter is probably stabilizing the compilation time, but not the result of run-time expressions.  
 I use CPython 2.7.3 of the following.  
 In the second example, the expression  "strin" + "g"  is compiled at the time of compilation, and replaced with  "string"  is. Earlier two examples behave the same.  
 If we examine the bytecodes, we will 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)   
 Third example includes a run-time consensus As a result, it 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 Binari_Add 25 STORE_FAST 3 (S3) 28 LOAD_CONST 0 (none) 31 Return_Wall   
 > gt; ; & Gt; & Gt; S3a = "strin"> gt; & Gt; & Gt; S3 = s3a + "g"> gt; & Gt; & Gt; S3 "string" false and gt; & Gt; & Gt; Intern (S3) is "string" true

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