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    <p>A list in Scheme is either the empty list <code>()</code> (also known as <code>nil</code> in some Lisps), or a cons cell whose <code>car</code> (also known as <code>first</code>) is an element of the list and whose <code>cdr</code> (also known as <code>rest</code>) is either the rest of the list (i.e., another list), or an atom that terminates the list. The conventional terminator is the empty list <code>()</code>; lists terminated by <code>()</code> are said to be "proper lists". Lists terminated by any other atom are called "improper lists". The list <code>(1 2 3 4 5)</code> contains the elements 1, 2, 3, 4, and 5, and is terminated by <code>()</code>. You could construct it by </p> <pre><code>(cons 1 (cons 2 (cons 3 (cons 4 (cons 5 ()))))) </code></pre> <p>Now, when the system <em>prints</em> a cons cell, the general case is to print it by </p> <pre><code>(car . cdr) </code></pre> <p>For instance, the result of <code>(cons 1 2)</code> is printed as </p> <pre><code>(1 . 2) </code></pre> <p>Since lists are built of cons cells, you can use this notation for lists too:</p> <pre><code>'(1 2 3 4 5) == '(1 . (2 . (3 . (4 . (5 . ()))))) </code></pre> <p>That's rather clunky, though, so most lisps (all that I know of) have a special case for printing cons cells: if the <code>cdr</code> is a list (either another cons cell, or <code>()</code>), then don't print the <code>.</code>, and don't print the surrounding parenthesis of the <code>cdr</code> (which it would otherwise have, since it's a list). So, if you're seeing a result like </p> <pre><code>(1 2 3 . 4) </code></pre> <p>it means you've got an improper list that is terminated by the atom <code>4</code>. It has the structure</p> <pre><code>(1 . (2 . (3 . 4))) </code></pre> <p>Now the question is: where in your code did the list construction go awry? <code>..</code> is always supposed to return a proper list, so let's look at the cases: The first case always returns a proper list (the empty list):</p> <pre><code>((&gt; (add1 start) stop) (quote ())) </code></pre> <p>The second case looks like it can return something that's not a list (assuming that <code>(sub1 stop) == (- stop 1)</code>):</p> <pre><code>((eq? (add1 start) stop) (sub1 stop)) </code></pre> <p>Now, if <code>..</code> were functioning correctly, then the third case would always be returning a proper list (since <code>(cons x y)</code> is a proper list if <code>y</code> is):</p> <pre><code>(else (cons start (.. (add1 start) stop))) </code></pre> <p>Make your second case return a list and you should be all set.</p>
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    1. USJoshua Taylor
    1. USJoshua Taylor
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    1. COThank you for your advice - the explanation was clear and easy to understand.
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