Capturing Value instead of Reference in Lambdas [duplicate]

Question

I was slightly surprised by this example given by Eli Bendersky (http://eli.thegreenplace.net/2015/the-scope-of-index-variables-in-pythons-for-loops/)

>>> def foo(): ... lst = [] ... for i in range(4): ... lst.append(lambda: i) ... print([f() for f in lst]) ... >>> foo() [3, 3, 3, 3]

But when I thought about it, it made some sense — the lambda is capturing a reference to i rather than i's value.

So a way to get around this is the following:

>>> def foo(): ... lst = [] ... for i in range(4): ... lst.append((lambda a: lambda: a)(i)) ... print([f() for f in lst]) ... >>> foo() [0, 1, 2, 3]

It appears that the reason that this works is that when i is provided to the outer lambda, the outer lambda creates a scope and dereferences i, setting a to i. Then, the inner lambda, which is returned, holds a reference to a.

Is this a correct explanation?

Worth reading: stackoverflow.com/questions/20246523/… and stackoverflow.com/a/14414638/1252759 — Jon Clements
– Jon Clements, Commented Jan 18, 2015 at 21:46

bav · Accepted Answer · 2015-01-18 21:45:38Z

15

Default param is an another way to catch a value:

lst.append(lambda i=i: i)

answered Jan 18, 2015 at 21:45

bav

1,63315 silver badges13 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

7stud · Accepted Answer · 2015-01-18 23:01:01Z

It appears that the reason that this works is that when i is provided to the outer lambda, the outer lambda creates a scope and dereferences i, setting a to i. Then, the inner lambda, which is returned, holds a reference to a.

Is this a correct explanation?

I don't like it. Python does not pass by reference:

def func(x): x = 10 num = 3 func(num) print num #=>3

As a result, the terms reference and dereference are not in the python lexicon. Or, you could say that python always dereferences a function argument before assigning it to a parameter variable--so your explanation doesn't really explain anything.

The reason the example works is because of the rule:

A function's local variables are destroyed after it finishes executing.

A function's local variables include its parameter variables. Every time the outer lambda executes, a new 'a' variable is created. As a result, each inner lambda closes over a different 'a' variable.

You did allude to that state of affairs:

the outer lambda creates a scope

...

the lambda is capturing a reference to i rather than i's value.

Or, as I like to phrase it.

A closure closes over variables--not values.

That is the way closures work in most languages(an exception being perl, where closures close over values).

Community · Accepted Answer · 2017-05-23 12:15:29Z

Yes, it looks correct. If you are familiar with javascript and know closures, you will notice how similar they are.

If not - there is a nice explanation on SO regarding JS closures and the concept is absolutely the same (as well as the explanation and even wrong and correct usage).

Collectives™ on Stack Overflow

Capturing Value instead of Reference in Lambdas [duplicate]

3 Answers 3

Comments

Comments

1 Comment

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

1 Comment

Linked

Related