comments.md

A comment on https://nedbatchelder.com/text/names.html

Myth: Python assigns mutable and immutable values differently.

While it is true that Python's assignment semantics are the same for both mutable and immutable types, the results look different for augmented assignments. E.g. the code

a = [1, 2, 3]
b = a
a += 4,

creates a new list, points both a and b to that list, and then appends another element. After this snippet ran, there is still only a single list, pointed to by both a and b. If you do the same with tuples, though

a = 1, 2, 3
b = a
a += 4,

you end up with two tuples: a is (1, 2, 3, 4), while b is still (1, 2, 3).

So augmented assignments behave differently for lists and tuples. This difference is not caused by different code being generated in both cases, but rather by a difference in the implementations of list.__iadd__() and tuple.__iadd__(). The line a += 4, generates code that is equivalent to

a = type(a).__iadd__(a, (4,))

for both the first and the second example. The implementation of __iadd__() for a list modifies the list in place and then returns a reference to the list that was passed in, so the assignment does not change the value a points to. For tuples, on the other hand, __iadd__() creates and returns a new tuple, so the augmented assignments results in an actual reassignment of the left-hand side.

Things get really crazy if you try augmented assignments of mutable elements in a tuple:

>>> a = [],
>>> a[0] += [1]
Traceback (most recent call last):
  File "<ipython-input-13-ea29ca190a4d>", line 1, in <module>
    a[0] += [1]
TypeError: 'tuple' object does not support item assignment

>>> a
([1],)

This code appends to a list embedded in a tuple, which should be possible, since a list is mutable. And the augmented assignment indeed modifies the list in place, but you still get an exception. If you look at the equivalent code the augmented assignment gets translated to, it's easy to understand why:

a[0] = type(a[0]).__iadd__(a[0], [1])

Names are local to the current scope, by default

Something that I find very interesting (and which can also cause bugs) is that python scans an entire function to look for name rebinding before determining the scope of a name.

Consider this REPL session:

>>> name = "Alice"
>>> 
>>> def foo():
...     print("Hello " + name)
... 
>>> def bar():
...     print("Hello " + name)
...     if False:
...         name = None
... 

At first glance, you would probably guess that these two functions should produce the same output. (For example, even if the False were changed to True, a C++ version of this code would still print "Hello Alice" for both functions, though/since name rebinding/scoping is explicit in C++.)

But here's what happens in Python:

>>> foo()
Hello Alice
>>> bar()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in bar
UnboundLocalError: local variable 'name' referenced before assignment

Huh, I had no idea a += b could do something different from a = a + b. In ruby a += b is just shorthand for a = a + b.

Slightly off-topic, but I am not a fan of overloading operators or special methods. I once spent several hours debugging some code in edx-platform where there was an if-else statement that would use a different branch than what I expected, and it was only once I realized that AccessResponse overrides __nonzero__, which can make an AccessResponse object evaluate to False in if statements.

I find that modifying the boolean behaviour of the class is much more confusing and doesn't really add anything over simply implementing a has_access() method or something similar.

@mtyaka So in Ruby, when you have two arrays a and b, the statement a += b results in the two original arrays being unchanged, and a pointing to a new array? That's certainly weird as well. The statement definitely reads as if a is changed in place.