Using a function as function parameter

Question

I tried to write a nested function to calculate the area. However, the program showed the name circle is not defined.

Could anyone please help providing a suggestion about this nested function problem? What concept is missing from the below code?

def area(shape, n):
    def circle(radius):
        return 3.14*radius**2
    def square(length):
        return length*length

print(area(circle, 10)   #system return -> name 'circle' is not defined
print(area(square, 5)   #system return -> name 'square' is not defined

What concept is missing from the below code? - Binding of names, Resolution of names, Python Scopes and Namespaces. — wwii
– wwii, Commented Jun 2, 2022 at 16:04
The concept you're missing is that everything defined inside a function, including nested functions, are local to their container. — martineau
– martineau, Commented Jun 2, 2022 at 16:11
Some good Q&A's searching with python functions in a class or module site:stackoverflow.com or python function grouping site:stackoverflow.com or something similar. — wwii
– wwii, Commented Jun 2, 2022 at 16:32

wwii · Accepted Answer · 2022-06-02 16:28:14Z

2

Instead of encapsulating those in a function, put them in a module.

area.py...

def circle(radius):
    return 3.14*radius**2
def square(length):
    return length*length

Then use them by importing area :

other.py

import area
q = area.circle(10)
z = area.square(5)

another.py

import area
shape = 'circle'
q = getattr(area,shape)(10)
shape = 'square'
z = getattr(area,shape)(5)

and.py

from area import circle,square
q = circle(10)
z = square(5)

edited Jun 2, 2022 at 16:28

answered Jun 2, 2022 at 15:58

wwii

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1 Comment

Mechanic Pig Over a year ago

It seems that circle should not be implemented as a nested function. I suspect that OP has a spelling error.

Santiago Ivulich · Accepted Answer · 2022-06-02 16:02:25Z

0

Circle is only defined in the scope of the area funcion. Functions are just like any other variable, you can read more here.

def circle(radius):
    return 3.14*radius**2

def square(length):
    return length*length

def area(shape, n):
    return shape(n)

print(area(circle, 10)
print(area(square, 5)

answered Jun 2, 2022 at 16:02

Santiago Ivulich

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Comments

LJustin · Accepted Answer · 2022-06-02 16:09:56Z

0

Since The functions are nested inside of area. Circle and square won't be recognised outside of the area function. I don't know if you need them to be nested. But you can just put them outside of the area function.

def circle(radius):
    return 3.14*radius**2
def square(length):
        return length*length

def area(shape, n):
    if shape == circle:
        print(circle(n))  

    if shape == square:
        print(square(n))  

area(circle, 20)
area(square, 20)

There are a lot of ways to do this.

def area(shape, n):
    if shape == "circle":
        return 3.14*n**2 

    if shape == "square":
        return n*n 

print(area("circle", 20))
print(area("square", 20))

answered Jun 2, 2022 at 16:09

LJustin

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Comments

quamrana · Accepted Answer · 2022-06-02 16:11:12Z

0

You can have nested functions, but you can't access them unless they are, for example, returned from that inner scope:

def area():
    def circle(radius):
        return 3.14*radius**2
    def square(length):
        return length*length
    return circle,square

circle, square = area()

print(circle(10))
print(square(5))

Output:

314.0
24

answered Jun 2, 2022 at 16:11

quamrana

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