Chained Assignment in Python Bytecode
20 points by wenderen 5 days ago | 15 comments

jp57 8 hours ago
The author's expectations seem strange. Take another example:

    a = b = random.random()
I would not expect a and b to get different values. It would be very strange if using `[]` had different behavior than a function call in the same place. Am I out of step here?
reply
maxnoe 4 hours ago
But this exact example is different, because the critical step of mutation is missing.

The initial line is the same, but:

    a = b = random.random()
    a += 1
    a == b # False
Only because floats are immutable and thus an implicit copy is made and lists are mutable so the same mutable instance is pointed to by both names.

This talk still applies despite its age: https://youtu.be/_AEJHKGk9ns?si=q5HjMOM9QS3_bFzH

reply
jibal 6 hours ago
What expectations? The author states right up front "I've known of this behavior for a long time".

A somewhat trickier example of the same issue is using [] as a default parameter value ... though there are warnings about the problem with that (it's the same list on every call) throughout the documentation.

reply
KerrickStaley 6 hours ago

  a = b = []
has the same semantics here as

  b = []
  a = b
which I don't find surprising.
reply
selridge 4 hours ago
“Since Python 3.6, every Python instruction has been given an odd number of arguments (even if it doesn't need any) so that the byte offsets are always even.”

I don’t think this is true. I think something like it is true for CPython (an argument isn’t added but the byte is kept for functions w no arg) but it’s not a statement about Python.

reply
Twirrim 6 hours ago
I've seen stuff posted about chained assignment footguns in python regularly over the years, and it always surprises me. I don't think I've ever written them, or reviewed code that does. I don't think it'd occur to me to even think about writing a chained assignment.

Is chained assignment a pattern that comes from another language that people are applying to python?

reply
kccqzy 9 hours ago
Chained assignments are banned according to the style guide at my workplace. Too many opportunities for misuse. And if you insist on a one-liner assignment to two variables just use two statements separated by the semicolon. I challenge anyone to work out what this code does:

    a, b = b[a] = 1, [0, 1, 2, 3]
reply
PurpleRamen 7 hours ago
> I challenge anyone to work out what this code does

It's unusual, but pretty obvious. In single steps it's basically this:

    a, b = 1, [0, 1, 2, 3]
    b[a] = b 
which would be b[1]=b because a==1. So this creates a self referencing list. The deeper reason here is, everything is a pointer to data in memory, even if in source code we see the actual data. That's why b[1] is storing the pointer to the list, not the data of the list.

If someone is doing BS like this, they deserve spanking. But banning the whole concept because people are unaware or how something is to be used properly is strange. But then again, it seems people have a bit of a problem how python really works and how it's different from other languages.

reply
js2 6 hours ago
Basically, but not quite. :-) The original result for b is:

  0, (1, [...]), 2, 3]
vs your version:

  [0, [...], 2, 3]
The equivalent statements to the original chained assignment are:

  a, b = 1, [0, 1, 2, 3]
  b[a] = 1, b  # relies on cpython implementation detail¹

¹: Using 1 works because the integers -5 thru 256 are interned in cpython. Otherwise, or if you don't want to rely on an implementation detail, to be truly equivalent (i.e. `id(b[1][0]) == id(a)`), it's this:

  a, b = 1, [0, 1, 2, 3]
  b[a] = a, b
reply
kccqzy 7 hours ago
An adjacent thread has some confusion about whether chained assignments happen left to right or right to left. Honestly that’s a factoid I don’t expect most Python programmers to know. It’s usually a bad idea to rely on people knowing arcade details of a language, especially a language like Python that has attracted many non-programmers like data scientists. (I have nothing against data scientists but their brainpower shouldn’t be wasted on remembering these kind of details.)
reply
js2 7 hours ago
I've been programming Python since 1.5.2 days and indeed, I didn't know the order of evaluation of chained assignments.

That said, it's the self-referencing list in your example that's the more confusing part. It's atypical to have self-referencing data structures, so that's something I'd comment in the design if I needed one.

reply
mathisfun123 9 hours ago
> list object is constructed once and assigned to both variables

Ummm no the list is constructed once and assigned to b and then b is assigned to a. It would be crazy semantics if `a = b = ...` meant `a` was assigned `...`.

Edit: I'm wrong it's left to right not right to left, which makes the complaint in the article even dumber.

reply
kccqzy 8 hours ago
It’s assigned left to right, not right to left. It’s documented in the Python language reference.

> An assignment statement evaluates the expression list and assigns the single resulting object to each of the target lists, from left to right.

Consider this:

    a = [1, 2]
    i = a[i] = 1
If assignment were to happen right to left, you would get a NameError exception because the first assignment would require an unbound variable.
reply
mathisfun123 7 hours ago
Fine but that even moreso illustrates how goofy the expectation that the "ctor" for [] would be called twice.
reply
ayhanfuat 8 hours ago
> then b is assigned to a

Wouldn't that require a LOAD_FAST? Also a is assigned first (from left to right) so a = ... happens either way.

reply