Python a, b = b, a +b

Let's say we start with a and b like this:

a = 2
b = 3

So, when you do:

a, b = b, a + b

what happens is you create the tuple (b, a + b) or (3, 5) and then unpack it into a and b so a becomes 3 and b becomes 5.

In your second example:

a = b
# a is now 3
b = a + b
# b is 3 + 3, or 6.

The line:

a, b = b, a + b

is closer to:

temp_a = a
a = b
b = temp_a + b

where b is using the old value of a before a was reassigned to the value of b.

Python first evaluates the right-hand expression and stores the results on the stack, then takes those two values and assigns them to a and b. That means that a + b is calculated before a is changed.

See How does swapping of members in the python tuples (a,b)=(b,a) work internally? for the low-down on how this all works, at the bytecode level.

Let's grok it.

a, b = b, a + b

It's a tuple assignment, means (a, b) = (b, a + b), just like (a, b) = (b, a)

Start from a quick example:

a, b = 0, 1
#equivalent to
(a, b) = (0, 1)
#implement as
a = 0
b = 1

When comes to (a, b) = (b, a + b)
EAFP, have a try directly

a, b = 0, 1
a = b #a=b=1
b = a + b #b=1+1
#output
In [87]: a
Out[87]: 1
In [88]: b
Out[88]: 2

However,

In [93]: a, b = b, a+b
In [94]: a
Out[94]: 3
In [95]: b
Out[95]: 5

The result is different from the first try.

Tha's because Python firstly evaluates the right-hand a+b
So it equivalent to:

old_a = a
old_b = b
c = old_a + old_b
a = old_b
b = c

In summary, a, b = b, a+b means,
a exchanges to get old_value of b,
b exchanges to get the sum of old value a and old value b,

In a, b = b, a + b, the expressions on the right hand side are evaluated before being assigned to the left hand side. So it is equivalent to:

c = a + b
a = b
b = c

In the second example, the value of a has already been changed by the time b = a + b is run. Hence, the result is different.