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POWithin Scala, is it possible to alias a type but disallow cross-use of aliased/non-aliased types like Haskell?
 

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  1. POWithin Scala, is it possible to alias a type but disallow cross-use of aliased/non-aliased types like Haskell?
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    <p>In Haskell, I believe that it is possible to alias a type in such a way that the compiler does not allow references between the aliased type and the unaliased type. According to <a href="https://stackoverflow.com/questions/2649305/why-is-there-data-and-newtype-in-haskell">this stack overflow question</a>, one can use Haskell's <code>newtype</code> like so:</p> <pre><code>newtype Feet = Feet Double newtype Cm = Cm Double </code></pre> <p>where <code>Feet</code> and <code>Cm</code> will behave like Double values, but attempting to multiply a <code>Feet</code> value and a <code>Cm</code> value will result in a compiler error.</p> <p>EDIT: Ben pointed out in the comments that this above definition in Haskell is insufficient. <code>Feet</code> and <code>Cm</code> will be new types, on which there will be no functions defined. Doing a bit more research, I found that the following will work:</p> <pre><code>newtype Feet = Feet Double deriving (Num) newtype Cm = Cm Double deriving (Num) </code></pre> <p>This creates a new type that derives from the existing <code>Num</code> type (requires using switch: <code>-XGeneralizedNewtypeDeriving</code>). Of course, these new types will be even more valuable deriving from other types such as <code>Show</code>, <code>Eq</code>, etc. but this is the minimum required to correctly evaluate <code>Cm 7 * Cm 9</code>.</p> <p>Both Haskell and Scala have <code>type</code>, which simply aliases an existing type and allows nonsensical code such as this example in Scala:</p> <pre><code>type Feet = Double type Cm = Double val widthInFeet: Feet = 1.0 val widthInCm: Cm = 30.48 val nonsense = widthInFeet * widthInCm def getWidthInFeet: Feet = widthInCm </code></pre> <p>Does Scala have a <code>newtype</code> equivalent, assuming that this does what I think it does?</p>
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    1. COThat's not actually how `newtype` works in Haskell. It creates a **new** type, which by default has no functions defined on it. So `Feet` and `Cm` values in your examples will not be able to be multiplied at all until you implement multiplication for them. Types declared with `newtype` will be *represented* identically to the wrapped type, which means that there is zero runtime cost of implementing operations on the `newtype` by simply unwrapping and passing through to the operation on the wrapped type. But that's really an optimization, irrelevant to what `newtype` actually means.
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    2. COThanks for the information, I updated the question to reflect that. Still a Haskell noobie, though very interested in getting better. =)
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    3. COYou should note that, while adding Cm to Cm to get Cm makes sense, multiplying Cm by Cm would get you Cm², which measures area rather than length. Using a Haskell-style type system to keep track of units can be pretty tricky to do properly.
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