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Today's Topics:
1. Is my understanding of ">>=" correct? (Martin Drautzburg)
2. Re: Is my understanding of ">>=" correct? (Bob Ippolito)
----------------------------------------------------------------------
Message: 1
Date: Tue, 19 Feb 2013 20:13:21 +0100
From: Martin Drautzburg <martin.drautzb...@web.de>
Subject: [Haskell-beginners] Is my understanding of ">>=" correct?
To: beginners@haskell.org
Message-ID: <201302192013.21463.martin.drautzb...@web.de>
Content-Type: text/plain; charset="us-ascii"
Hello all,
It took me quite some time to understand >>=. What confused me, was that the
second operand is a function, not just a Monad. I believe I am beginning to
see the light.
First there is no way to sequence operations by just writing one after the
other. However we can chain operations by passing the output of one operation
to the next. But this does not not allow capturing the intermediate results.
With >>= the output of one operation gets magically bound to the parameter of
the function and remains in scope all the way through.
I was confused by the signature m a -> (a -> mb) -> mb because it looks like
there is (as second parameter) a function which returns a Monad. But I doubt I
ever saw such a function, in the sense that it does something intelligent. It
is typically a function that does not do a whole lot with its parameter, but
just returns another Monad, such that its parameter is bound to the result of
the previous expression.
But it could be done:
do
x <- SomeMonad
y <- someCleverlyConstructedMonad x
If I want to see the function of the >>= signature I have to read the do block
right->left (following the arrow) and then diagonally to the lower right. The
diagonal steps reveal the function. In the above example we have
\x -> someCleverlyConstructedMonad
but many times I just see
do
x <- SomeMonad
y <- someOtherMonad
and the function does not do anything intelligent but just binds x and
enforces the sequence of operations.
is this about right?
--
Martin
------------------------------
Message: 2
Date: Tue, 19 Feb 2013 12:15:54 -0800
From: Bob Ippolito <b...@redivi.com>
Subject: Re: [Haskell-beginners] Is my understanding of ">>=" correct?
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Message-ID:
<cacwmpm_bzz3bsl40im9b_rusxbrp8j3hvmyxhzoq3utxteo...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
On Tue, Feb 19, 2013 at 11:13 AM, Martin Drautzburg <
martin.drautzb...@web.de> wrote:
> Hello all,
>
> It took me quite some time to understand >>=. What confused me, was that
> the
> second operand is a function, not just a Monad. I believe I am beginning to
> see the light.
>
> First there is no way to sequence operations by just writing one after the
> other. However we can chain operations by passing the output of one
> operation
> to the next. But this does not not allow capturing the intermediate
> results.
> With >>= the output of one operation gets magically bound to the parameter
> of
> the function and remains in scope all the way through.
>
Well, sequencing actions is a special case of chaining actions where you
simply ignore the result of the previous action. This is the >> operator.
It is defined as follows:
m >> k = m >>= \_ -> k
There is no magic involved, just functions and lexically scoped variables.
I was confused by the signature m a -> (a -> mb) -> mb because it looks like
> there is (as second parameter) a function which returns a Monad. But I
> doubt I
> ever saw such a function, in the sense that it does something intelligent.
> It
> is typically a function that does not do a whole lot with its parameter,
> but
> just returns another Monad, such that its parameter is bound to the result
> of
> the previous expression.
>
I think it's important to note that the signature is a -> (a -> m b) -> m
b. The space is significant. 'm' is a separate type variable from 'a' or
'b'.
> But it could be done:
>
> do
> x <- SomeMonad
> y <- someCleverlyConstructedMonad x
>
> If I want to see the function of the >>= signature I have to read the do
> block
> right->left (following the arrow) and then diagonally to the lower right.
> The
> diagonal steps reveal the function. In the above example we have
>
> \x -> someCleverlyConstructedMonad
>
> but many times I just see
>
> do
> x <- SomeMonad
> y <- someOtherMonad
>
> and the function does not do anything intelligent but just binds x and
> enforces the sequence of operations.
>
> is this about right?
>
It's not really possible to translate the above example into valid code
because a do block must end with an expression. Let's say you were to
change it to the following valid syntax:
do
x <- firstAction
y <- secondAction x
return (x, y)
The de-sugaring of that do block would look like this (newlines added to
show function scope):
firstAction >>= \x ->
secondAction x >>= \y ->
return (x, y)
A concrete example:
do
x <- getLine
y <- print x
return (x, y)
h> getLine >>= \x -> print x >>= \y -> return (x, y)
Hello!
"Hello!"
("Hello!",())
do blocks are just syntax sugar, they don't do anything you couldn't easily
do with 'let ... in', '>>=' and '->'. They just make it a lot easier to
read in some cases. Other cases it's a wash, such as in this example where
we don't want to capture any intermediate results:
do
x <- getLine
print x
is equivalent to:
getLine >>= print
-bob
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