python - Explain this inconsistency -

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• in python, why can function modify arguments perceived caller, not others? 8 answers

here 2 methods. 1 modifies variable x, other not. can please explain me why 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 local variable.

x = [0, 0, 0, 0] 

reassigns local variable x new list. not affect global variable x of same name.

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you remove x arguments of swatch:

def swatch():     x = [0, 0, 0, 0] 

but when python encounters assignment inside function definition like

x = [0, 0, 0, 0] 

python consider x local variable default. assigning values x not affect global variable, x.

to tell python wish x global variable, need use global declaration:

def swatch():     global x     x = [0,0,0,0] swatch() 

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however, in case, since x mutable, define swatch this:

def swatch(x):     x[:] = [0,0,0,0] 

although x inside swatch local variable, since swatch called

swatch(x)  # global variable x 

it points same list global variable of same name.

x[:] = ... alters contents of x, while x still points original list. thus, alters value global variable x points too.

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def switch(a,b,x):      x[a], x[b] = x[b], x[a] 

is example contents of x mutated while x still points original list. mutating local x alters global x well.