« Tuple Unpacking Oddness

Aleksa Sarai

programming python

01 September 2015

I just wanted to start out by saying that I’m not entirely sure what the standards people where smoking when they stated that the following is entirely expected behaviour. I also want to thank my fellow tutors for helping me get to grips with what is going on here.

The Code

The code in question looks pretty innocuous:

a = "a" b = "b" L = [a, b] # Here it comes ... L[L.index(a)], L[L.index(b)] = b, a

Now, what do you expect the above code to do? I personally would expect (with my experiences with tuple unpacking) that L would be [b, a] . However, it seems that Python has other ideas:

>>> # Insert code from above. >>> L == [b, a] False >>> L == [a, b] True

… Wait, what? So not only didn’t it do what we’d expect (that the values would be switched), but in addition it didn’t even affect the list? Even though clearly the first portion of the tuple simply must modify the list, right?

However, if we switch the order of the list, we see what we’d expect:

>>> L = [b, a] >>> L[L.index(a)], L[L.index(b)] = b, a >>> L == [b, a] False >>> L == [a, b] True

So what the hell is going on here? It looks like the list is always going to end up being [a, b] using this method. There’s clearly something fishy going on here.

Can’t Touch dis 1

So, after staring at the code for ten minutes or so (and some very generous help from my fellow tutors) we nailed what the problem was. First we busted out dis to see what CPython was actually executing and in what order.

>>> # Insert the code from above in a function called `func`. >>> import dis >>> dis.dis(func) # SNIP: Setup instructions. 7 24 LOAD_FAST 1 (b) 27 LOAD_FAST 0 (a) 30 ROT_TWO 31 LOAD_FAST 2 (L) 34 LOAD_FAST 2 (L) 37 LOAD_ATTR 0 (index) 40 LOAD_FAST 0 (a) 43 CALL_FUNCTION 1 (1 positional, 0 keyword pair) 46 STORE_SUBSCR 47 LOAD_FAST 2 (L) 50 LOAD_FAST 2 (L) 53 LOAD_ATTR 0 (index) 56 LOAD_FAST 1 (b) 59 CALL_FUNCTION 1 (1 positional, 0 keyword pair) 62 STORE_SUBSCR # SNIP: Return instructions.

The key points here are the positions of the STORE_SUBSCR instructions in relation to the calls to .index . As you can see, Python decides to modify the list L before evaluating the subscripts. Extracting only the important lines:

# SNIP: Loading code. 43 CALL_FUNCTION 1 (1 positional, 0 keyword pair) # .index(a) 46 STORE_SUBSCR # b # SNIP: Loading code. 59 CALL_FUNCTION 1 (1 positional, 0 keyword pair) # .index(b) 62 STORE_SUBSCR # a

So, what CPython has decided to generate is less like this (which is what we might expect):

# Temporary variables for the indexes (to ensure "correct" order of operations). ia = L.index(a) ib = L.index(b) # Expanded tuple. L[ia] = b L[ib] = a

And in fact generates something vaguely similar to this:

L[L.index(a)] = b L[L.index(b)] = a

Which explains why there’s no change! We modify L[0] and then revert it immediately. This seemed to be (at least in my opinion) a violation of the order of operations of subscripting and tuple unpacking.

1 I’m sorry, I’m so sorry. I just couldn’t resist.

Versions of Python Affected

This appeared to affect the following Python versions and implementations:

CPython 3.4.3

CPython 2.7.10

PyPy 2.6.0

Which lead me to believe that this probably isn’t some implementation-specific, hacky bytecode optimisation being done by CPython (which was my first impression when I was looking at the bytecode dis spat out). It must be documented somewhere.

The Standard

So, as with all weirdness, in the event of confusion it is best to refer to the specification. I want to thank David Vo for linking me the relevant section of the standard (I don’t like trawling through standards docs to find the one sentence that perfectly defines your situation).

In particular, §7.2 states that:

If the target list is a comma-separated list of targets: […] the items are assigned, from left to right, to the corresponding targets. […] Assignment of an object to a single target is recursively defined as follows […] if the target is a subscription: The primary expression in the reference is evaluated. It should yield either a mutable sequence object […] Next, the subscript expression is evaluated. [emphasis added]

And, in fact, it references a very similar example to ours in the docs:

Although the definition of assignment implies that overlaps between the left-hand side and the right-hand side are ‘simultanenous’ (for example a, b = b, a swaps two variables), overlaps within the collection of assigned-to variables occur left-to-right, sometimes resulting in confusion.

So, it does seem that this behaviour is very well defined (and is completely expected). However, it is something to watch out for as it definitely can cause confusion (as it did me). Panic over.

Happy Pythoning. And remember, there’s no such thing as too much magic.

Unless otherwise stated, all of the opinions in the above post are solely my own and do not necessary represent the views of anyone else. This post is released under the Creative Commons BY-SA 4.0 license.

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