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<p class="MsoNormal">Hello – in</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal" style="text-indent:36.0pt">myFoldl :: (t -> a -> t) -> t -> [a] -> t<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">the t and the a are type variables. There are no “classes” at all. Since you didn’t name the type variables, Haskell has provided default names for you “at random”, and it’s in some ways a little unfortunate that it’s named one of them
t.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">You could (probably should) provide your own declaration, e.g. renaming t to b:<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal" style="text-indent:36.0pt">myFoldl :: (b -> a -> b) -> b -> [a] -> b<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">This represents exactly the same thing, and says that myFoldl will work for any two types a and b. It takes three params:</p>
<p class="MsoNormal"> (b -> a -> b) A function of values of type b and type a</p>
<p class="MsoNormal" style="margin-left:72.0pt;text-indent:36.0pt">that returns a value of type b. (The accumulation function)</p>
<p class="MsoNormal"> b A value of type b (The initial accumulated value)</p>
<p class="MsoNormal"> [a] A list of values of type a (This list to accumulate over)</p>
<p class="MsoNormal">and returns</p>
<p class="MsoNormal"> b A value of type b (The final accumulated value)</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">The prelude function:</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal" style="text-indent:36.0pt">foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">is then almost identical. This time, though, instead of being restricted to a list of values (of type a) to fold over, it can fold over any type of foldable collection of values (of type a). t represents the type of the foldable collection,
and Foldable t => is a constraint that says the function will only work for a type t that is an instance of the class Foldable.</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">[<i>BTW: </i><i>does a type declaration like this one reflect in any way the recursion going on?</i> I didn’t really understand the question, but you can often deduce what a function does, and how it might be implemented, from its type
signature.]<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">I don’t know how you’re going about learning Haskell, but following a tutorial (such as
<a href="http://learnyouahaskell.com/">http://learnyouahaskell.com/</a>) might be helpful, and I think explains all of the above step by step.</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Regards, David.</p>
<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal" style="border:none;padding:0cm"><b>From: </b><a href="mailto:borgauf@gmail.com">Lawrence Bottorff</a><br>
<b>Sent: </b>02 January 2021 17:52<br>
<b>To: </b><a href="mailto:beginners@haskell.org">The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell</a><br>
<b>Subject: </b>[Haskell-beginners] Type explanation of foldl?</p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">I've got this<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">myFoldl f init [] = init<br>
myFoldl f init (x:xs) = myFoldl f newInit xs<br>
where newInit = f init x<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">which when evaluated gives this for its type<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">> :t myFoldl<o:p></o:p></p>
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<p class="MsoNormal">myFoldl :: (t -> a -> t) -> t -> [a] -> t<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">. . . not totally sure what that all means, e.g., how the t as a generic class is behaving is very mysterious. Obviously, the final result is of type t . . . and that must relate to the incoming argument init, correct? (BTW, does a type
declaration like this one reflect in any way the recursion going on?) But then with Prelude foldl I'm really clueless<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">> :t foldl<o:p></o:p></p>
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<p class="MsoNormal">foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">I'm guessing the t is again a generic type of the Foldable class, but then. . . Can someone walk me through this?<o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">LB<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
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