Seriously, what's the difference? When you first fire up the Python interpreter, __builtins__
is in your namespace for free:
>>> globals().keys()
['__builtins__', '__name__', '__doc__', '__package__']
>>> __builtins__
<module '__builtin__' (built-in)>
But it appears to be the __builtin__
module (singular)! If you:
>>> import __builtin__
>>> __builtin__ is __builtins__
True
Hrm. So they are both names that point to the same object, the module __builtin__
. Weird. Why does Python do this? Do they always behave the same?
I read on StackOverflow that
By default, when in the
__main__
module,__builtins__
is the built-in module__builtin__
(note: no 's'); when in any other module,__builtins__
is an alias for the dictionary of the__builtin__
module itself.
What. What does that mean.
This talk of the "__main__
module" and "any other module" reminds me of a sequence of words that I've known for quite a while, but haven't completely grokked:
We can access the name of the current module with the builtin variable
__name__
.
You're probably familiar with the related canonical statement:
if __name__ == '__main__':
main()
But what does "current module" mean? What does the __name__
variable look like when it does not equal __main__
?
I happen to know, because I've obsessively read about the import
statement, another sequence of words:
Any code executed as a result of an
import
isn't executed in the__main__
module.
Let's use these bits of knowledge to observe the behavior of __builtins__
both inside and outside of the __main__
module. We can also check out the __name__
variable while we're at it.
First, let's make a script, a.py
, which will allow us to observe the behavior of __builtin__
, __builtins__
, and __name__
.
import __builtin__
print "In a"
print "__name__ is:", __name__
print "__builtin__ is __builtins__:", __builtin__ is __builtins__
print "type(__builtin__):", type(__builtin__)
print "type(__builtins__):", type(__builtins__)
Let's see what happens when we execute a.py
:
$ python a.py
In a
__name__ is: __main__
__builtin__ is __builtins__: True
type(__builtin__): <type 'module'>
type(__builtins__): <type 'module'>
Okay. So we're in the __main__
module, and in here __builtin__
is pointing to the same module object as __builtins__
.
What happens if we import a
in another script? The code in a
will execute, but it won't be executed within the __main__
module. Instead, it'll be executed within the a
module. Let's write another script, b.py
, to find out what happens to __builtins__
outside of __main__
:
import __builtin__
print "In b, before importing a"
# the output from this should be the same as when we ran
# $ python a.py
print "__name__ is:", __name__
print "__builtin__ is __builtins__:", __builtin__ is __builtins__
print "type(__builtin__):", type(__builtin__)
print "type(__builtins__):", type(__builtins__)
print "\n"
import a
# code from a will execute here
Let's see what happens when we run b.py
:
$ python b.py
In b, before importing a
__name__ is: __main__
__builtin__ is __builtins__: True
type(__builtin__): <type 'module'>
type(__builtins__): <type 'module'>
In a
__name__ is: a
__builtin__ is __builtins__: False
type(__builtin__): <type 'module'>
type(__builtins__): <type 'dict'>
Aha. So when we're outside the context of the __main__
module, __name__
is just equal to the name of the module where code is currently being executed. That seems logical. And outside of __main__
, __builtins__
is a dict, rather than a module.
We were told earlier that, outside the context of __main__
, "__builtins__
is an alias for the dictionary of the __builtin__
module". I think that means that __builtins__ is __builtin__.__dict__
. Let's see if my hypothesis is true, by adding another line to the bottom of our a.py
file:
print "__builtins__ is __builtin__.__dict__", __builtins__ is __builtin__.__dict__
Running b.py
again, we get:
$ python b.py
In b, before importing a
__name__ is: __main__
__builtin__ is __builtins__: True
type(__builtin__): <type 'module'>
type(__builtins__): <type 'module'>
In a
__name__ is: a
__builtin__ is __builtins__: False
type(__builtin__): <type 'module'>
type(__builtins__): <type 'dict'>
__builtins__ is __builtin__.__dict__ True
Yes! My hypothesis was correct. Okay. So now I get why using __builtin__
is better than __builtins__
:
The type, and thus behavior, of __builtins__
changes based on the context of where it's being executed, while the type and behavior of __builtin__
is constant. Rad.
Thanks, stranger who suggested I look into this, for the learning opportunity. And thanks, always, to Allison Kaptur, for exploring this topic with me.
The code for this blog post is on GitHub, of course.
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