Reputation: 61
First I'm going to start like everyone else. I'm new to python. My teacher gave me the problem:
def f(a, b, c):
a = 1
c = b
c[0] = 2
a = 10
b = [11, 12, 13]
c = [13, 14, 15]
f(a, b, c)
print a, b, c
It prints:
10 [2, 12, 13] [13, 14, 15]
I understand that a stays at 10 because integers are immutable, but I don't understand why b changes and c doesn't.
Upvotes: 6
Views: 512
Reputation: 54242
c = b
c[0] = 2
Since you're setting c
to point to b
, You could just as easily do this:
def f(a, b, unused): # notice c isn't in the parameter list
a = 1
c = b # c is declared here
c[0] = 2 # c points to b, so c[0] is b[0]
Now it's obvious that c
is always the same as b
, so why not just remove it:
def f(a, b, unused):
a = 1
b[0] = 2
And now it's clear that you're changing the first element of b
and not doing anything to c
, and remember, this is functionally identical to the original.
Upvotes: 4
Reputation: 47988
a
does not retain the value of 10 because it is immutable. It retains the value of 10 because when you call a = 1
in the local scope of f()
you create a new variable.
When you call c = b
inside f()
, the local c
becomes a local reference to the mutable object represented by b
. When you re-assign the values in that mutable object the change is reflected in the original object.
Upvotes: 1
Reputation: 5714
The key is to understand the variables as pointers under the hood:
def f(a, b, c):
a = 1 # a is a single scalar value, so no pointing involved
c = b # point the local "c" pointer to point to "b"
c[0] = 2 # change the 2nd value in the list pointed to by "c" to 2
When you call f(a,b,c), only b actually gets changed. The "c" variable inside the function implementation is different from the "c" implementation outside the function.
Upvotes: 2