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Casting: (NewType) vs. Object as NewType

By : user78841
Date : July 29 2020, 05:00 AM
Hope this helps The former will throw an exception if the source type can't be cast to the target type. The latter will result in sc2 being a null reference, but no exception.
code :

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Does D have 'newtype'?

By : Xavier Normand
Date : March 29 2020, 07:55 AM
wish of those help In D1.0 there is typedef, which is the strong typing from a predefined type to a 'newtype.'
D2.0 has removed this and only alias remains (what typedef is in C). There is talk about having a wrapper template that can strongly create a new type.
code :
typedef int Fish;
typedef Fish Cat;
Fish gold = 1;
Cat fluff = gold;
meters = cast(meters) 12.7;

Haskell unsafeCoerce newtype to identical newtype

By : swrg2012
Date : March 29 2020, 07:55 AM
Hope this helps This might be safe with the current GHC implementation, but this is not the recommended way to solve your particular problem.
The pattern that is usually used instead is to have an internal module like this:
code :
module Foo.Types (T(..)) where

newtype T = T Int
module Foo.Bla where

import Foo.Types (T(..))

f :: T -> Bla
f (T int) = makeBla int
module Foo (T) where

import Foo.Types (T(..))

makeT :: Int -> T
makeT = T

What is the WHNF of a newtype and how does rseq work on a newtype?

By : Sheila Snell
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Per the section on Datatype renamings in the report,
code :
Sum ⊥ = ⊥
Sum (lengthyComputation :: Int) `using` rseq
let x = Sum (lengthyComputation :: Int) `using` rseq
seq ⊥ b  =  ⊥
seq a b  =  b, if a ≠ ⊥
seq (Sum ⊥) b = ⊥
seq (lengthyComputaton :: Int) b
newtype Foo a = Foo { unFoo :: a }  -- record syntax for convenience below
function :: Foo a -> Bar
function (Foo x) = whatever x
function y = let x = unFoo y in whatever x

Function expects structure of newtype but not newtype itself

By : 程二狗
Date : March 29 2020, 07:55 AM
Any of those help I seem to miss something fundamental. ,
The type of map is
code :
map :: (a -> b) -> [a] -> [b]
join :: (Start, [End]) -> [Edge]
map join :: [(Start, [End])] -> [[Edge]]
type Node = Integer
type Edge = (Node, Node)
type Start = Node
type End = Node

newtype Graph = Graph {
  edges :: [(Start, [End])]
} deriving (Eq, Show)

join :: (Start, [End]) -> [Edge]
join (start, ends) = map (\e -> if start < e then (start, e) else (e, start)) ends

sampleGraph =
  Graph [(5, [1, 2, 3]),
         (7, [1, 2]),
         (1, [1, 2]),
         (2, [1, 2])

foo = map join . edges
case sampleGraph of
   Graph edges -> map join edges

Converting from newtype to Int and from Int to newtype

By : Neena Antony
Date : March 29 2020, 07:55 AM
around this issue Use fromIntegral to convert between integral types, such as Int and Integer. Use NT1 to convert from Integer to NT1, and pattern-matching from NT1 to Integer. Finally, you can compose functions to connect things.
code :
toNT1 :: Int -> NT1
toNT1 = NT1 . fromIntegral

fromNT1 :: NT1 -> Int
fromNT1 (NT1 x) = fromIntegral x
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