Chris Angelico <ros...@gmail.com>:
> Okay. Now let's suppose that, instead of "73" in the first step, you
> have "ask the user for an integer". Are you allowed to eliminate this
> prompt, since the result of it cannot possibly affect anything? And if
> not, why not?
I would guess yes; that's how Python works as well:
>>> 7 or input()
7
However, if we think of the I/O interaction (aka *the process*) to be
the return value of a function, every bead in the I/O chain counts.
> After all, eliminating terms that have been multiplied by zero is one
> form of algebra.
I wonder if the word "algebra" brings anything to this discussion. It
doesn't make Haskell any more or less functional.
> If you consider that the world changes state as a result of asking the
> user for input, then you've just eliminated all notion of functional
> purity.
Not necessarily. Nothing changes the world. Rather you have different
worlds: the world of the past and the world of the future. The world of
the past is in the past of the world of the future:
def print(world, text, cont):
return cont(World(past=world, offset=text))
def print_x_then_y(world, x, y, cont):
return print(world, x, lambda world2: print(world2, y, cont))
> You have side effects, plain and simple, and you're using imperative
> code.
Technically, no. Thought-model-wise, yes.
Of course, recursive functions can simulate Turing machines and vice
versa. You can write purely functional code in purely imperative style.
My example above is purely functional as:
* Every object is immutable.
* The order of evaluation does not change the end result.
The end result is an ordered sequence of events. The topological order
is a causal order (you need the past world to construct the future
world), and causal order generates a temporal order.
Now, *if* the Haskell VM chooses to *realize* I/O events *as soon as* it
becomes aware of them, you get traditional, imperative side effects.
Marko
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