CODEWITHSUNDEEP
pythonpython-3.x
gpa
gpa

Reputation: 3

Defining a function with another function

Could someone help explain what's going on here? I don't understand what is going on in the third part, and why the result is 9. Thank you!

>>> def square(x):
        return x ** 2

>>> def f(x):
        return x * x

>>> def try_f(f):
        return f(3)

>>> try_f(square)
9

Upvotes: 0

Views: 541

Answers (3)

Andriy Ivaneyko
Andriy Ivaneyko

Reputation: 22021

you call square function by parameter passing from try_f function and pass 3 as argument to it. You can add print to observe which function is called.

Defining f function doesn't take affect try_f behavour

Upvotes: 0

Tiger-222
Tiger-222

Reputation: 7150

When calling try_f(square) you are passing the square function to try_f. Inside try_f you named the first argument f: it will have nothing to do with the f() function defined below. This is now a local variable to the current scope of try_f.

As a better example, take this:

def square(x):
    return x * x

def double(x):
    return x * 2

def try_f(func):
    return func(4)

>>> try_f(square)
16
>>> try_f(double)
8

Upvotes: 1

hectorcanto
hectorcanto

Reputation: 2266

The third function has a function as a parameter so when called it runs the parameter-function with a 3 as a paramater.

try_f(f=square) resolves as square(x=3) which resolves as x*x= 3*3 = 9

Upvotes: 0

Related Questions