Hard coded variables in python function

Question

Sometimes, some values/strings are hard-coded in functions. For example in the following function, I define a "constant" comparing string and check against it.

def foo(s):
    c_string = "hello"
    if s == c_string:
        return True
    return False

Without discussing too much about why it's bad to do this, and how it should be defined in the outer scope, I'm wondering what happens behind the scenes when it is defined this way.
Does the string get created each call?
If instead of the string "hello" it was the list: [1,2,3] (or a list with mutable content if it matters) would the same happen?

Community · Accepted Answer · 2017-05-23 11:53:23Z

11

Because the string is immutable (as would a tuple), it is stored with the bytecode object for the function. It is loaded by a very simple and fast index lookup. This is actually faster than a global lookup.

You can see this in a disassembly of the bytecode, using the dis.dis() function:

>>> import dis
>>> def foo(s):
...     c_string = "hello"
...     if s == c_string:
...         return True
...     return False
... 
>>> dis.dis(foo)
  2           0 LOAD_CONST               1 ('hello')
              3 STORE_FAST               1 (c_string)

  3           6 LOAD_FAST                0 (s)
              9 LOAD_FAST                1 (c_string)
             12 COMPARE_OP               2 (==)
             15 POP_JUMP_IF_FALSE       22

  4          18 LOAD_GLOBAL              0 (True)
             21 RETURN_VALUE        

  5     >>   22 LOAD_GLOBAL              1 (False)
             25 RETURN_VALUE        
>>> foo.__code__.co_consts
(None, 'hello')

The LOAD_CONST opcode loads the string object from the co_costs array that is part of the code object for the function; the reference is pushed to the top of the stack. The STORE_FAST opcode takes the reference from the top of the stack and stores it in the locals array, again a very simple and fast operation.

For mutable literals ({..}, [..]) special opcodes build the object, with the contents still treated as constants as much as possible (more complex structures just follow the same building blocks):

>>> def bar(): return ['spam', 'eggs']
... 
>>> dis.dis(bar)
  1           0 LOAD_CONST               1 ('spam')
              3 LOAD_CONST               2 ('eggs')
              6 BUILD_LIST               2
              9 RETURN_VALUE

The BUILD_LIST call creates the new list object, using two constant string objects.

Interesting fact: If you used a list object for a membership test (something in ['option1', 'option2', 'option3'] Python knows the list object will never be mutated and will convert it to a tuple for you at compile time (a so-called peephole optimisation). The same applies to a set literal, which is converted to a frozenset() object, but only in Python 3.2 and newer. See Tuple or list when using 'in' in an 'if' clause?

Note that your sample function is using booleans rather verbosely; you could just have used:

def foo(s):
    c_string = "hello"
    return s == c_string

for the exact same result, avoiding the LOAD_GLOBAL calls in Python 2 (Python 3 made True and False keywords so the values can also be stored as constants).

edited May 23, 2017 at 11:53

CommunityBot

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answered Aug 24, 2015 at 10:09

Martijn Pieters

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3 Comments

jonrsharpe Over a year ago

Doesn't the same thing happen with s == "hello"?

Arthur.V Over a year ago

You are right about the comment on tuples, I should've asked about list instead. Ill correct the question. The rest of the answer is excellent for now :)

Martijn Pieters Over a year ago

@ArthurVaiselbuh: I covered lists now too, as well as {...} dictionaries and sets.

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