Ross Paterson wrote:
> On Mon, Jan 19, 2009 at 01:13:37PM -0800, Jonathan Cast wrote:
>
>> (On the other hand, your hunch that lift = return is correct --- so you
>> get a cookie for that; it's just that return here is neither the return
>> of the monad for m nor the return of the monad for Read
G'day all.
Quoting Jonathan Cast :
(By the way, you *do* have the equations
lift (return x) = return x
[...]
Right. And you could, at least in principle, implement "return" this
way in all monad transformers.
Cheers,
Andrew Bromage
___
Haskel
On Mon, 2009-01-19 at 21:31 +, Ross Paterson wrote:
> On Mon, Jan 19, 2009 at 01:13:37PM -0800, Jonathan Cast wrote:
> > (On the other hand, your hunch that lift = return is correct --- so you
> > get a cookie for that; it's just that return here is neither the return
> > of the monad for m nor
On Mon, Jan 19, 2009 at 01:13:37PM -0800, Jonathan Cast wrote:
> (On the other hand, your hunch that lift = return is correct --- so you
> get a cookie for that; it's just that return here is neither the return
> of the monad for m nor the return of the monad for ReaderT m. It is,
> instead, the r
On Mon, 2009-01-19 at 13:03 -0800, Daryoush Mehrtash wrote:
> Is there a reason why the lift function in ReaderT's MonadTrans instance is
> implemented as:
>
> instance MonadTrans (ReaderT r) where
>
> lift m = ReaderT $ \_ -> m
>
>
>
> Instead of just using the monad's return function?
On Mon, Jan 19, 2009 at 01:03:55PM -0800, Daryoush Mehrtash wrote:
> lift m = ReaderT $ \_ -> m
> return a = ReaderT $ \_ -> return a
If you look carefully you will see that these are not the same.
-Brent
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Haskell
Is there a reason why the lift function in ReaderT's MonadTrans
instance is implemented as:
instance MonadTrans (ReaderT r) where
lift m = ReaderT $ \_ -> m
Instead of just using the monad's return function? Could "lift m"
be implemented as "return m"?
instance (Monad m) => Monad (Rea
