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POType error while trying to implement the (>>=) function in order to create a custom monad transformer
 

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  1. POType error while trying to implement the (>>=) function in order to create a custom monad transformer
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    <p>I'm trying to create a monad transformer for a future project, but unfortunately, my implementation of the Monad typeclasse's (>>=) function doesn't work.</p> <p>First of all, here is the underlying monad's implementation : </p> <pre><code>newtype Runtime a = R { unR :: State EInfo a } deriving (Monad) </code></pre> <p>Here, the implementation of the Monad typeclasse is done automatically by GHC (using the <code>GeneralizedNewtypeDeriving</code> language pragma). The monad transformer is defined as so : </p> <pre><code>newtype RuntimeT m a = RuntimeT { runRuntimeT :: m (Runtime a) } </code></pre> <p>The problem comes from the way I instanciate the (>>=) function of the Monad typeclasse :</p> <pre><code>instance (Monad m) =&gt; Monad (RuntimeT m) where return a = RuntimeT $ (return . return) a x &gt;&gt;= f = runRuntimeT x &gt;&gt;= id &gt;&gt;= f </code></pre> <p>The way I see it, the first <code>&gt;&gt;=</code> runs in the underlying <code>m</code> monad. Thus, <code>runRuntimeT x &gt;&gt;=</code> returns a value of type <code>Runtime a</code> (right ?). Then, the following code, <code>id &gt;&gt;=</code>, should return a value of type <code>a</code>. This value is the passed on to the function f of type <code>f :: (Monad m) =&gt; a -&gt; RuntimeT m b</code>. </p> <p>And here comes the type problem : the <code>f</code> function's type doesn't match the type required by the (>>=) function. Jow can I make this coherent ? I can see why this doesn't work, but I can't manage to turn it into something functionnal.</p> <p>Edit : The error message : </p> <pre><code>Core.hs:34:4: Occurs check: cannot construct the infinite type: m = RuntimeT m When generalising the type(s) for `&gt;&gt;=' In the instance declaration for `Monad (RuntimeT m)' Failed, modules loaded: none. </code></pre> <p>Thank you for you help, and do not hesitate to correct any flaws in my message,<br> Charlie P.</p>
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