Is it possible to call a nested function defined inside an existing function:
For example:
def first(x):
def second():
print(x)
return second
I know I can do something like this: first(10)()
Yet I want to do something similar to this:
first(10).second()
My thinking is that it is not possible because second does not exist until first is called.
Am I right?
If you want, you can return several functions that way:
class Box(object):
def __init__(self,**kw): vars(self).update(kw)
def first(x):
def second():
print(x)
def third(y): return x+y
return Box(second=second,third=third)
first(10).second() # prints 10
x=first(15)
x.second() # prints 15
print(x.third(10)) # prints 25
Any resemblance to reputation scores for answers is coincidental.
first to be a class? It seems to be doing the same thing in a less idiomatic way.
first was a method; changing it to a class would require passing self around explicitly and wreak havoc with code organization and/or indentation. This technique also lets you naturally have a set of "methods" that is variable. I'm not claiming it should be anyone's first choice.
Why not use a class?
class First:
def __init__(self, x):
self.x = x
def second(self):
print(x)
First(3).second()
The fact that all Python functions are closures is a great feature, but it's fairly advanced. Classes are the usual place to store state.
Though it is not the right way but just for fun. What you are doing here is returning the instance of the second function, so while calling it, no dont need to call it by name, jsut use the variable.
def first(x):
def second():
print(x)
return second
x = first(10)
x()
or like this
first(10)()
as mention by other, use class instead.
seconddoesn't exist outsidefirstscope. Consider using a class insteadsecond(), but why not makefirst()a class to begin with? (Example:def first(x): return str(x), thenfirst(10).upper()would work, because strings have a method namedupper()).