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POHaskell Type Coercion
 

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  1. POHaskell Type Coercion
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    <p>I trying to wrap my head around Haskell type coercion. Meaning, when does can one pass a value into a function without casting and how that works. Here is a specific example, but I am looking for a more general explanation I can use going forward to try and understand what is going on:</p> <pre><code>Prelude&gt; 3 * 20 / 4 15.0 Prelude&gt; let c = 20 Prelude&gt; :t c c :: Integer Prelude&gt; 3 * c / 4 &lt;interactive&gt;:94:7: No instance for (Fractional Integer) arising from a use of `/' Possible fix: add an instance declaration for (Fractional Integer) In the expression: 3 * c / 4 In an equation for `it': it = 3 * c / 4 </code></pre> <p>The type of (/) is Fractional a => a -> a -> a. So, I'm guessing that when I do "3 * 20" using literals, Haskell somehow assumes that the result of that expression is a Fractional. However, when a variable is used, it's type is predefined to be Integer based on the assignment.</p> <p>My first question is how to fix this. Do I need to cast the expression or convert it somehow? My second question is that this seems really weird to me that you can't do basic math without having to worry so much about int/float types. I mean there's an obvious way to convert automatically between these, why am I forced to think about this and deal with it? Am I doing something wrong to begin with?</p> <p>I am basically looking for a way to easily write simple arithmetic expressions without having to worry about the neaty greaty details and keeping the code nice and clean. In most top-level languages the compiler works for me -- not the other way around. </p>
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    1. COQuick note: Try doing this in a source file, not in GHCi directly. It will figure more things out for you that way, because it can look at the whole program, not deal with a line at a time.
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    2. COThis question is based on a whole program that is not working. I've isolated this specific portion that the compiler is complaining about. I can post the entire thing, but I don't think it will help much.
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    3. COYour question's been answered, but: "there's an obvious way to convert automatically between these." Is there really? What should `2^128 :: Integer` automatically become when I want a `Double`, or even an `Int64`? And of course, going from non-integral types to integral types can't be automatic because of rounding. In general, Haskell doesn't have subtyping or automatic type coercions; it has (constrained) parametric polymorphism. But once a type has become concrete/monomorphic, values of that type are of *only* that type.
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      1. USAntal Spector-Zabusky
 

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