
I see misunderstand comes from
flg <- p x
In fact in the following code flg has type Bool:
flg <- [True, False]
For lists, >>= (and therefore <-) assigns every value out of list to
flg and tries to proceed. The final `return` returns only a single
element out of result, and all the results are concatenated together.
On Wed, Apr 22, 2015 at 12:22 PM, Shishir Srivastava
Hi Mike,
Thanks for your response. I was aware that 'casting' doesn't really fit in Haskell vocab but for the lack of better word used it.
My question was however more towards the usage of [Bool] in the 'if' statement of the filterM function.
More precisely - How does 'if [True, False] then x else y' work , because when I do this in GHCi it throws up the following error ?
Couldn't match expected type `Bool' with actual type `[Bool]'
Clearly the 'if' construct does not take a list of Boolean but a single Boolean value so how does filterM use it in it's implementation.
Hope have made myself clear this time.
Thanks, Shishir
From: Mike Meyer
To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Cc: Date: Wed, 22 Apr 2015 04:15:31 -0500 Subject: Re: [Haskell-beginners] filterM function On Wed, Apr 22, 2015 at 3:31 AM, Shishir Srivastava
wrote: I still don't quite understand how 'flg' being a boolean [] is used in the last 'if statement' of implementation because when I try to do the same thing outside in GHCi it fails miserably even though I am casting it to [Int] -
-- return (if [True, False] then "4" else "3")::[Int]
"cast" is a misnomer in Haskell. When you add a type to an expression, you aren't changing the type of the expression like a C-style cast, but picking out a type from the set of possible types for that expression.
Ignoring the if part and and focusing on return, which has a type Monad m => a -> m a. [Int] is equivalent to [] Int, so [] would be the Monad, and a is Int. While 3 can be an Int, "3", can't. So you could do return 3 :: [Int] or equivalently return 3 :: [] Int to get [3], you can't do return "3" :: [Int], because "3" can't be an Int. You can do return "3" :: [String], since "3" is a string. Just to show the range of possibilities, you can do return 3 :: IO Float, and get back 3.0 as an IO action. The monad in the type is IO, and 3 can be interpreted as a Float.
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