Comma separated variable assignment [duplicate]

Answer 1

For the purposes of reading, all it's doing is setting a and b, so that a = 0 and b = 1. Similarly, in line 7, it's setting a to b and b to the sum of a and b.

More specifically, it's setting tuples. Tuples are invariant, in that once they're created, their values can't change. Tuples are pervasive in python - you see them almost everywhere.

Typically, you would expect a tuple to be in parenthesis, e.g. (a, b) = (0, 1) would read more cleanly, but they are such a large feature of python that the parenthesis are optional (unless you're constructing a tuple as an argument to a function, and then you need the extra parenthesis to differentiate between a single tuple and multiple arguments. I.e. you would have to say foo((a, b)) to pass a tuple to foo, as foo(a, b) would pass two arguments to it.)

Tuples can be any length. You can write a, b, c, d, e = 0, 1, 2, 3, 4, or you can have a function return a tuple, e.g.: ret1, ret2, ret3 = foobar(1)

Answer 2

What you describe is tuple assignment:

a, b = 0, 1

is equivalent to a = 0 and b = 1.

It can however have interesting effects if you for instance want to swap values. Like:

a,b = b,a

will first construct a tuple (b,a) and then untuple it and assign it to a and b. This is thus not equivalent to:

#not equal to
a = b
b = a

but to (using a temporary):

t = a
a = b
b = t

In general if you have a comma-separated list of variables left of the assignment operator and an expression that generates a tuple, the tuple is unpacked and stored in the values. So:

t = (1,'a',None)
a,b,c = t

will assign 1 to a, 'a' to b and None to c. Note that this is not syntactical sugar: the compiler does not look whether the number of variables on the left is the same as the length of the tuple on the right, so you can return tuples from functions, etc. and unpack them in separate variables.