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    <p>As for the first one, it's a very basic way of recursion. However, it seems to be missing a part:</p> <pre><code>myLength [] = 0 </code></pre> <p>It works by scaling off one element at the time from the list and adding one to the result. To visualise, consider the call</p> <pre><code>myLength [1,2,3] </code></pre> <p>which will evaluate to:</p> <pre><code>1 + myLength [2,3] 1 + 1 + myLength [3] 1 + 1 + 1 + myLength [] 1 + 1 + 1 + 0 </code></pre> <p>which is 3.</p> <p>As for the second one, well, you have already split the string at the next line break into two parts: pre and suf. Now, suf will start with either a \n, or a \r, or a \r\n. We want to remove these. So we use pattern matching. See how the rest variable is essentially the suf variable minus the initial line break character(s).</p> <p>So we have pre, which is the first line, and rest, which is the rest of the text. So in order to continue splitting rest into lines we call splitLines on it recursively and concatenate the result to pre.</p> <p>To visualize, say you have the string "foo\nbar\r\nbaz".</p> <p>So, when calling, the result will be:</p> <pre><code>[ pre =&gt; foo, suf =&gt; \nbar\r\nbaz, rest =&gt; bar\r\n\baz ] foo : splitLines bar\r\nbaz </code></pre> <p>then splitLines is called again, and the result is expanded into:</p> <pre><code>[ pre =&gt; bar, suf =&gt; \r\nbaz, rest = baz ] foo : bar : splitLines baz </code></pre> <p>then once again:</p> <pre><code>[ pre =&gt; baz, suf =&gt; [], rest = [] ] foo : bar : baz </code></pre> <p>which is the final result.</p>
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    1. POCould someone explain these Haskell functions to me?
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    1. POCould someone explain these Haskell functions to me?
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    1. COI understand how this works, but now I'm lost as to why this works. The result is either another recursive call, or an empty list. Essentially, it looks like pre is getting concatenated to pre every recur, but the case expression isn't returning pre. Confusing.
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    2. COActually after reading it again, it looks like pre is getting concatenated to the argument to splitLines, not the result. I'm just not sure how this is returning anything more than a recursive call which will eventually return []. I guess I don't understand recursion as much as I thought I did :\
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    3. COWhen I look at the function, the case expression looks like it's returning the result of a recursive call which is the result of a recursive call which is the result of a recursive call, and so on until it returns []. I don't see how rest is being concatenated to pre in this situation.
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