CODEWITHSUNDEEP
python
Sophie
Sophie

Reputation: 25

how is this lambda function retrieving its x value?

I don't understand how this lambda function knows x is equal to 1?

def one(f = None): return 1 if not f else f(1)
def two(f = None): return 2 if not f else f(2)
def plus(y): return lambda x: x+y

one(plus(two()))
3

I know that the inner function two() returns 2 because f is defaulted to None. Thus y= 2. But how does the lambda function know to look to the outmost function for the x value?

Upvotes: 0

Views: 161

Answers (3)

U13-Forward
U13-Forward

Reputation: 71570

Let's run through the steps first:

  1. one gets a value, f, as plus(two()).

  2. plus gets a value as two, two is gonna be 2 since no f.

  3. Okay, so since one gets the value of that, it gonna condition and see that it shouldn't return 1, so does f(1) the f is the unfinished lambda process, the lambda needs one parameter more to add up, so it got 1, so 2 + 1 is 3.

That is the whole process really.

Upvotes: 0

J. Kwon
J. Kwon

Reputation: 174

If you look at one() it passes "1" into a function which you pass in the arguments (if a function is passed. otherwise 1 is returned). Thus, it evaluates to f(1) (see in the else of one). The function you pass to one() is lambda x: x + 2 (since y=2). Thus, this evaluates to lambda 1: 1 + 2

If you call one(lambda x: 50), it returns 50.

Upvotes: 2

zvone
zvone

Reputation: 19352

plus returns a (lambda) function. That function is passed to one. Within the scope of one, it is called f.

Then f (which is actually the lambda from returned from plus) is called in f(1).

In other words, the code one(plus(two())) does this:

number2 = two()
lambda_function = plus(number2)
result = one(lambda_function)

Upvotes: 2

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