At what moment are closure cells evaluated? Can lambas have closures?

Question

Consider the following code:

def A():
    l = list()
    for i in range(5):
        l.append(lambda: i)
    return l

for f in A():
    print( f() )

It prints out 4 five times, and I'm assuming it does that because the lambda function has taken the variable i through the outer scope of A, but right before the moment the variable became out of scope.

Then we have this code:

def B():
    l = list()
    for i in range(5):
        l.append((lambda i: lambda: i)(i))
    return l

for f in B():
    print( f() )

It prints out all numbers from 0 to 4 in order. I am assuming it does that because the variable i has been grabbed from the scope of the outer lambda which took it as a parameter, so the value has been assigned to the cell in the moment when that external lambda finished executing.

But what if i held a mutable object instead of an immutable int?

def C():
    l, i = list(), list()
    for r in range(5):
        i.append(r)
        l.append((lambda i: lambda: i)(i.copy()))
    return l

for f in C():
    print( f() )

It does print out the lists in order as expected, because I used the list.copy() method in the argument.

Now, is my understanding correct?

If so, then is there a more pythonic or simpler way to save in a closure cell either an immutable object (or a copy of a mutable object), at the exact moment I want to? In other words, is there a better way than the (lambda i: lambda: i)(i) latch I came up with?

Does this answer your question? [Python lambda's binding to local values](https://stackoverflow.com/questions/10452770/python-lambdas-binding-to-local-values) — juuso, Jan 29 '22 at 11:47
Closures capture variables, not values. A closure doesn't save the state of a variable at some particular time; it saves the variable itself. — user2357112, Jan 29 '22 at 12:09

juuso · Answer 1 · 2022-01-29T16:20:47.080

This happens because of late binding, try this code:

def bind(value):
  return lambda: value

def A():
    l = list()
    for i in range(5):
        l.append(bind(i))
    return l

for f in A():
    print( f() )

This way you can obtain the expected behaviour, since the binding is now on a value and not on a variable. Of course this is something tricky, and you should pay attemption to the behaviours of function like this, since python is using late binding.

At what moment are closure cells evaluated? Can lambas have closures?

1 Answers1