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copied!<ol> <li><p>This does not mean what you think:</p> <pre><code>fun funkyPlus 3 = (fn x => fn y => x*x*y)3 </code></pre> <p>It defines a function which takes an argument that must be 3, and which evaluates to its RHS if it is 3 and is undefined otherwise. What you mean to say is this: If we only provide an argument for x, we have the following:</p> <pre><code>funkyPlus 3 → (fn x => fn y => x*x+y) 3 </code></pre> <p>and so forth.</p></li> <li><p>Secondly, there is an error in your <code>foldl</code>:</p> <pre><code>fun foldl f b [] = b|foldl f b (h::t) = foldl f f(h,b) t; ^^^^^ Type clash: expression of type 'a * 'b cannot have type 'c list </code></pre> <p>This is because <code>(h,b)</code> is parsed as the third argument to <code>foldl</code> and not as the argument to <code>f</code>. Parenthesize it:</p> <pre><code>fun foldl f b [] = b|foldl f b (h::t) = foldl f (f(h,b)) t; > val ('a, 'b) foldl = fn : ('a * 'b -> 'b) -> 'b -> 'a list -> 'b </code></pre></li> </ol> <p>Now, getting to your question, ML can tell us that an expression like <code>foldl add</code> would have type <code>int -> int list -> int</code>.</p> <p>But in general, it may help to realize that function application is entirely mechanical. If we have these two definitions:</p> <pre><code>fun foldl f b [] = b | foldl f b (h::t) = foldl f (f(h,b)) t; add (x,y) = x + y; </code></pre> <p>then <code>var example = foldl add</code> would be equivalent to this:</p> <pre><code>fun example b [] = b | example b (h::t) = example (h::t) (add(h,b)) t; </code></pre> <p>All that’s been done is that <code>add</code> has been substituted for <code>f</code> in the body of <code>foldl</code>, nothing more (although I have taken the liberty of replacing <code>foldl add</code> with <code>example</code> in the body).</p>

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