I have tried the following in the console:
>>> def f(l=[]):
... l.append(1)
... print(l)
... del l
...
>>> f()
[1]
>>> f()
[1, 1]
What I don't understand is how the interpreter is still able to find the same list l after the delete instruction.
From the documentation l=[] should be evaluated only once.
The variable is not the object. Each time the function is called, the local variable l is created and (if necessary) set to the default value.
The object [], which is the default value for l, is created when the function is defined, but the variable l is created each time the function runs.
list, del l[:] would do that; l would remain bound to the list (being unbound when the function returns immediately thereafter as normal), but the list itself would have its contents deleted so it's clear for the next call.
del for a default function argument since I did not find anything about that in the internet.To delete a element for the list,del l[:] should be used.If u use just l the list will remain itself.
def f(l=[]):
l.append(1)
print(l)
del l[:]
print(l)
>>> f()
[1] #element in the list
[] #list after deletion of the element
>>> f()
[1]
[]
>>> f()
[1]
[]
lis created and set to the value of the default argument. The default argument's value is created on function definition, but the variablelis created each time the function runs.delof a local variable, right at the end of a function, is NEVER going to have any effect whatsoever. It's exactly what's going to happen to the variable anyway.