It never makes sense to use foldl on lists in Haskell (2019)

Re:It never makes sense to use foldl on lists in Haskell (2019) by toolslive on Saturday March 21, 2020 @06:09AM (#22645601)

They are, but it's shallow:

    >>> x = ([1,2,3,4],"whatever")
    >>> x[0].append('z')
    >>> x
    ([1, 2, 3, 4, 'z'], 'whatever')

And python gives you access to anything:

    >>> def gotcha():
    ...      z = globals()['x']
    ...      z = list(z)
    ...      z.append("I've seen'em do it man!")
    ...      globals()['x'] = tuple(z)
    ... 
    >>> x = (1,2,3,4)
    >>> gotcha()
    >>> x
    (1, 2, 3, 4, "I've seen'em do it man!")

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Re:It never makes sense to use foldl on lists in Haskell (2019) by didibus on Saturday March 21, 2020 @05:41PM (#22650426)

It be shallow with a Singly Linked List too no? What inherently makes a Singly Linked List better at addressing this problem if it contains mutable elements?

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Re:It never makes sense to use foldl on lists in Haskell (2019) by toolslive on Sunday March 22, 2020 @05:39PM (#22658944)

Typically, in a functional language, values are immutable. So if you cons to a list, you create a new value that has the current list as its tail. Since all lists are immutable, they can be shared and there's no need for a deep copy.

Python is not a functional programming language....

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