问题
I have this snippet:
a = [1,2,3]
b = a
b = [4,5,6] # a doesnt change
and this:
a = [1,2,3]
b = a
b[0] = 5 # a also changes
How is b's initialization playing a part in deciding the mutability of a?
回答1:
When you create a list and assign it to a variable, like
a = [1, 2, 3]
You create an object, and the variable a holds a reference to that object. When you do
b = a
You assign the same reference in b, so now both a and b point to your list. So when you execute
b[0] = 5
You alter that same list.
You can see this in action by using the id() function, which returns the memory address of the object:
>>> a = [1, 2, 3]
>>> b = a
>>> id(a), id(b)
(140454191340720, 140454191340720)
They are identical. The point is, a and b are not lists themselves, they point to a list.
When you do something like:
a = [1, 2, 3]
b = a
b = [2, 3, 4]
You first assigned to b a reference to the list that a points to, but then you assign a new reference to it.
By the way, this can bite you in the behind when you do something along the lines of
def foo (a=[]):
a.append(42)
return a
since the argument a points to the same list on every invocation, if you call this function 5 times without arguments, the list will contain 5x 42:
foo(); foo(); foo(); foo();
print(foo())
>>> [42, 42, 42, 42]
来源：https://stackoverflow.com/questions/16731404/understanding-mutability-in-python