[Haskell-cafe] forever function laziness

[Andras Slemmer]

>> What is confusing to me is the recursion on the let construct.
Oh sry I may have misunderstood your question then.

You can think of defining 'forever' itself as a let construct. Starting
from the "simpler" implementation:

let forever a = a >> forever a

Now let's just give the rhs another name, again using 'let':

let forever a = (let a' = a >> forever a in a')

But we can see that a' is actually equal to (forever a), so we can replace
on in the rhs:

let forever a = (let a' = a >> a' in a')

There is no trickery, no getting away, this recursion is the same as what
you have thought of:)



On 24 December 2013 02:46, Eduardo Sato <eduardo.sato at gmail.com> wrote:

> On Tuesday, December 24, 2013, Clark Gaebel wrote:
>
>> forever :: Monad m => m a -> m b
>> forever a = a >> forever a
>>
>
> Thanks for your response. This is actually how I would have implemented
> it.
>
> What is confusing to me is the recursion on the let construct.
>
> What is not clear to me is when we can "get away" with doing that kind of
> thing. I mentioned the "tying the knot article" because it, too, uses
> recursion on a let construct.
>
> --
> Eduardo Sato
>
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