Ability to unfreeze collection after `.freeze()` method is called

[Nikita Grishko]

Long story short

Hi,

In some cases, I need to unfreeze my collection, make some mutation, and freeze it again. With the current API of the collection, there is no way to do that without collection recreation.

I can submit a PR with changes I want to see in the library, just let me know if my proposal is acceptable.

Expected behaviour

l = FrozenList()
l.append(1)
l.freeze()

l.unfreeze()
l.append(2)

Actual behaviour

l = FrozenList()
l.append(1)
l.freeze()

l = FrozenList()
l.append(1)
l.append(2)

Steps to reproduce

Described in the examples above.

Your environment

[Martijn Pieters]

With the current API of the collection, there is no way to do that without collection recreation.

That's deliberate. Once frozen, it it is an immutable collection, like a tuple or a frozenset. Other code can then rely on the sequence to never change; it can be used in concurrent code.

You can't create an unfreeze() method without breaking this, so is not something we'll add here.

Instead, create a new FrozenList() that is a copy of the old, frozen FrozenList() instance, and replace the reference. That way existing references to the frozen copy can complete their work, then later on grab a reference to the new object.