Multiple assignment and evaluation order in Python

What is the difference between the following Python expressions:

# First:

x,y = y,x+y

# Second:

x = y
y = x+y

First gives different results than Second.

e.g.,

First:

>>> x = 1
>>> y = 2
>>> x,y = y,x+y
>>> x
2
>>> y
3

Second:

>>> x = 1
>>> y = 2
>>> x = y
>>> y = x+y
>>> x
2
>>> y
4

y is 3 in First and 4 in Second

In an assignment statement, the right-hand side is always evaluated fully before doing the actual setting of variables. So,

x, y = y, x + y

evaluates y (let's call the result ham), evaluates x + y (call that spam), then sets x to ham and y to spam. I.e., it's like

ham = y
spam = x + y
x = ham
y = spam

By contrast,

x = y
y = x + y

sets x to y, then sets y to x (which == y) plus y, so it's equivalent to

x = y
y = y + y

It is explained in the docs in the section entitled "Evaluation order":

... while evaluating an assignment, the right-hand side is evaluated before the left-hand side.

The first expression:

1. Creates a temporary tuple with value y,x+y
2. Assigned in to another temporary tuple
3. Extract the tuple to variables x and y

The second statement is actually two expressions, without the tuple usage.

The surprise is, the first expression is actually:

temp=x
x=y
y=temp+y

You can learn more about the usage of comma in "Parenthesized forms".

I've recently started using Python and this "feature" baffled me. Although there are many answers given, I'll post my understanding anyway.

If I want to swap the values of two variables, in JavaScipt, I'd do the following:

var a = 0;
var b = 1;

var temp = a;
a = b;
b = temp;

I'd need a third variable to temporarily hold one of the values. A very straightforward swap wouldn't work, because both of the variables would end up with the same value.

var a = 0;
var b = 1;

a = b; // b = 1 => a = 1
b = a; // a = 1 => b = 1

_{Imagine having two different (red and blue) buckets and having two different liquids (water and oil) in them, respectively. Now, try to swap the buckets/liquids (water in blue, and oil in red bucket). You can't do it unless you have an extra bucket.}

Python deals with this with a "cleaner" way/solution: Tuple Assignment.

a = 0
b = 1

print(a, b) # 0 1

# temp = a
# a = b
# b = temp

a, b = b, a # values are swapped

print(a, b) # 1 0

I guess, this way Python is creating the "temp" variables automatically and we don't have to worry about them.

An observation regarding the left-hand side as well: the order of assignments is guaranteed to be the order of their appearance, in other words:

a, b = c, d

is equivalent functionally to precisely (besides t creation):

t = (c, d)
a = t[0] # done before 'b' assignment
b = t[1] # done after 'a' assignment

This matters in cases like object attribute assignment, e.g.:

class dummy:
    def __init__(self): self.x = 0

a = dummy(); a_save = a
a.x, a = 5, dummy()
print(a_save.x, a.x) # prints "5 0" because above is equivalent to "a = dummy(); a_save = a; t = (5, dummy()); a.x = t[0]; a = t[1]"

a = dummy(); a_save = a
a, a.x = dummy(), 5
print(a_save.x, a.x) # prints "0 5" because above is equivalent to "a = dummy(); a_save = a; t = (dummy(), 5); a = t[0]; a.x = t[1]"

This also implies that you can do things like object creation and access using one-liners, e.g.:

class dummy:
    def __init__(self): self.x = 0
# Create a = dummy() and assign 5 to a.x
a, a.x = dummy(), 5