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Beginner: Converting Types in Haskell


Beginner: Converting Types in Haskell

By : user2955114
Date : November 22 2020, 10:38 AM
around this issue digitToInt is something that already exists. It used to live in the Char module but now it lives in Data.Char, so we have to import Data.Char to use it.
code :
Prelude> import Data.Char
Prelude Data.Char> digitToInt '1'
1
Prelude Data.Char> map digitToInt ['1', '2', '3', '4']
[1,2,3,4]
alwaysSeven :: Char -> Int
The type signature for `alwaysSeven' lacks an accompanying binding
alwaysSeven :: Char -> Int
alwaysSeven x = 7


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Using Data.Heap in Haskell, or reading Haskell docs for a beginner

Using Data.Heap in Haskell, or reading Haskell docs for a beginner


By : Jyotish
Date : March 29 2020, 07:55 AM
will be helpful for those in need Usually data structure libraries in Haskell provides fromList functions to convert from a list to that structure. Data.Heap is no exception. But you'll get some crazy errors when trying to use it:
code :
Prelude Data.Heap> Data.Heap.fromList [1,2,5,7]

<interactive>:1:0:
    Ambiguous type variables `t', `pol' in the constraint:
      `HeapItem pol t'
        arising from a use of `fromList' at <interactive>:1:0-27
    Probable fix: add a type signature that fixes these type variable(s)

....
Prelude Data.Heap> Data.Heap.fromList [1,2,5,7] :: MinHeap Int
fromList [(1,()),(2,()),(5,()),(7,())]
Prelude Data.Heap> let heap = Data.Heap.fromList [1,2,5,7] :: MinHeap Int
Prelude Data.Heap> heap
fromList [(1,()),(2,()),(5,()),(7,())]
Prelude Data.Heap> Data.Heap.insert 3 heap
fromList [(1,()),(3,()),(2,()),(5,()),(7,())]
Prelude Data.Heap> heap
fromList [(1,()),(2,()),(5,()),(7,())]
Prelude Data.Heap> viewHead heap
Just 1
How to avoid converting among different 'string' types in haskell, using snapframework?

How to avoid converting among different 'string' types in haskell, using snapframework?


By : Krands
Date : March 29 2020, 07:55 AM
like below fixes the issue Snap automatically decodes the request and makes it available to you through the Request data type. It provides functions getRequest and withRequest for retrieving the request and a number of other accessor functions for getting various parts.
There are also convenience functions for common operations. To get a POST or GET parameter see getParam.
Converting between types in Haskell

Converting between types in Haskell


By : CriBurn
Date : March 29 2020, 07:55 AM
To fix this issue The problem is not that you can't convert a Double to an Int — round accomplishes that just fine — it's that you're trying to do division on an Int (padLength - length string). The error message is just telling you that Int is not an instance of Fractional, the typeclass for numbers that can be divided.
In general, you could use fromIntegral (padLength - length string) to turn it into a Double, but in this case, you can simply use integer division directly:
code :
center padLength string = padLength - (length string) `div` 2
Converting enumerations to int-like types in haskell

Converting enumerations to int-like types in haskell


By : user5487679
Date : March 29 2020, 07:55 AM
help you fix your problem Sometimes GHC's advice is bad, but in this case it's spot on. At the top of your file, put
code :
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
(yellow:red:green:blue:empty:_) = map Color [1..]
Converting types in Haskell

Converting types in Haskell


By : Robin Clement
Date : March 29 2020, 07:55 AM
wish helps you Your type signature is wrong. If you convert something you can't write it into the type signature. Only the last one is the return type. The others are parameter types. Follow these:
code :
fc::(Bool,[Char])->Integer
fc (x,y) = toInteger . ord . head $ y

fc::(Bool,[Char])->Int->Integer--
fc (x,y) n = if n == w then n else w
 where w = toInteger . ord . head $ y
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