Here are two methods. One modifies the variable x, the other does not. Can you please explain to me why this is?
x = [1,2,3,4]
def switch(a,b,x):
x[a], x[b] = x[b], x[a]
switch(0,1,x)
print(x)
[2,1,3,4]
def swatch(x):
x = [0,0,0,0]
swatch(x)
print(x)
[2,1,3,4]
The function definition
def swatch(x):
defines x to be a local variable.
x = [0, 0, 0, 0]
reassigns the local variable x to a new list.
This does not affect the global variable x of the same name.
You could remove x from the arguments of swatch:
def swatch():
x = [0, 0, 0, 0]
but when Python encounters an assignment inside a function definition like
x = [0, 0, 0, 0]
Python will consider x a local variable by default. Assigning values to this x will not affect the global variable, x.
To tell Python that you wish x to be the global variable, you would need to use the global declaration:
def swatch():
global x
x = [0,0,0,0]
swatch()
However, in this case, since x is mutable, you could define swatch like this:
def swatch(x):
x[:] = [0,0,0,0]
Although x inside swatch is a local variable, since swatch was called with
swatch(x) # the global variable x
it points to the same list as the global variable of the same name.
x[:] = ... alters the contents of x, while x still points to the original list. Thus, this alters the value that the global variable x points to too.
def switch(a,b,x):
x[a], x[b] = x[b], x[a]
is another example where the contents of x are mutated while x still points to the original list. Thus mutating the local x alters the global x as well.
