Question

Can you describe difference between two ways of string concatenation: simple __add__ operator and %s patterns?
I had some investigation in this question and found %s (in form without using parentheses) a little faster.

Also another question was appeared: why result of 'hell%s' % 'o' refers to another memory region than 'hell%s' % ('o',)?

There is some code example:

l = ['hello', 'hell' + 'o', 'hell%s' % 'o', 'hell%s' % ('o',)]
print [id(s) for s in l]

Result:

[34375618400, 34375618400, 34375618400, 34375626256]

P.S. I know about string interning 🙂

Answer 1

Here is a small exercise:

>>> def f1():
    'hello'


>>> def f2():
    'hel' 'lo'


>>> def f3():
    'hel' + 'lo'


>>> def f4():
    'hel%s' % 'lo'


>>> def f5():
    'hel%s' % ('lo',)


>>> for f in (f1, f2, f3, f4, f5):
    print(f.__name__)
    dis.dis(f)


f1
  1           0 LOAD_CONST               1 (None) 
              3 RETURN_VALUE         
f2
  1           0 LOAD_CONST               1 (None) 
              3 RETURN_VALUE         
f3
  2           0 LOAD_CONST               3 ('hello') 
              3 POP_TOP              
              4 LOAD_CONST               0 (None) 
              7 RETURN_VALUE         
f4
  2           0 LOAD_CONST               3 ('hello') 
              3 POP_TOP              
              4 LOAD_CONST               0 (None) 
              7 RETURN_VALUE         
f5
  2           0 LOAD_CONST               1 ('hel%s') 
              3 LOAD_CONST               3 (('lo',)) 
              6 BINARY_MODULO        
              7 POP_TOP              
              8 LOAD_CONST               0 (None) 
             11 RETURN_VALUE         

As you can see, all simple concatenations/formatting are done by compiler. The last function requires more complex formatting and therefore, I guess, is actually executed. Since all those object created at compilation time they all have the same id.