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copied!<p>So backticks turn a binary function into an infix operator, making</p> <pre><code>x `op` y </code></pre> <p>equivalent to</p> <pre><code>op x y </code></pre> <p>So <code>op</code> needs to be of type <code>a -> b -> c</code> where <code>x :: a</code> and <code>y :: b</code>.</p> <p>In your case, <code>op</code> was <code>mempty</code>, with the type <code>Monoid m => m</code>. But we know it to be of the form <code>a -> b -> c</code>, so substitute that and you get (this is no longer valid syntax) <code>Monoid (a -> b -> c) => a -> b -> c</code>, because we can substitute that <code>m</code> for anything as long as the constraint holds.</p> <p>Now we know (due to the instance declaration) that any function of the form <code>s -> t</code>, where <code>t</code> is a Monoid, is a Monoid itself, and we also know that <code>a -> b -> c</code> is really <code>a -> (b -> c)</code>, i.e. a function taking one argument and returning another function. So if we substitute <code>a</code> for <code>s</code> and <code>(b -> c)</code> for <code>t</code>, the we fulfill the Monoid instance, if <code>t</code> is a Monoid. Of course, <code>t</code> is <code>(b -> c)</code>, so we can apply the same Monoid instance again (with <code>s = b</code> and <code>t = c</code>), so if <code>c</code> is a Monoid, we're good.</p> <p>So what is <code>c</code>? The expression you had was</p> <pre><code>PQ.singleton `mempty` [start] </code></pre> <p>i.e.</p> <pre><code>mempty PQ.singleton [start] </code></pre> <p>The instance declaration for <code>Monoid (a -> b)</code> defines <code>mempty _ = mempty</code>, i.e. it's a function that ignores its argument and returns the empty element of the <code>b</code> Monoid. In other words, we can expand the call above to</p> <pre><code>mempty [start] </code></pre> <p>i.e. we ignore the argument and use mempty of the inner Monoid (which is <code>b -> c</code>). Then we repeat, ignoring the argument again:</p> <pre><code>mempty </code></pre> <p>So the expression you had is just equivalent to a single <code>mempty</code>, which has the type <code>Monoid c => c</code>, i.e. it can be any Monoid whatsoever.</p> <p>In your case, the larger expression deduces <code>c</code> to be a <code>PQ.MinPQueue</code>. And <code>MinPQueue</code> is a Monoid instance with <code>mempty</code> being the empty queue.</p> <p>This is how you end up with the result you're seeing.</p>

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