Re: type (++) = (<>)

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Re: type (++) = (<>)

Doug McIlroy
> What do you think of making (++) the same as (<>)

Before Haskell 98, ++ was an overloadable operator associated with
class MonadPlus. Why was that arrangement abolished?

Doug
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Re: type (++) = (<>)

Doug McIlroy
> What do you think of making (++) the same as (<>)

This seems to be a call for returning to the old situation in
which (++) was an operator of class MonadPlus. Why was that
abolished in Haskell 98?

Doug
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Re: type (++) = (<>)

Lana Black
On 03/07/17 17:29, Doug McIlroy wrote:
>> What do you think of making (++) the same as (<>)
>
> This seems to be a call for returning to the old situation in
> which (++) was an operator of class MonadPlus. Why was that
> abolished in Haskell 98?
>
> Doug

Using MonadPlus in this case is not the best idea. There are types that
can implement (++) but aren't a Monad. Data.Set is one example.
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