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    <p>That second <code>(list ...)</code> in your <code>else</code> clause is what's screwing you up. It's nesting each successive call deeper and deeper. The recursion will naturally create the list; you don't need to wrap it again.</p> <p>Try:</p> <pre><code>(define get-ivars (λ (ivars num) (if (null? ivars) '() (cons (list (car ivars) `(nth args ,num)) (get-ivars (cdr ivars) (+ num 1)))))) </code></pre> <p>Regarding the <code>get-ivars</code> caller code, the parentheses surrounding the unquoted call to <code>get-ivars</code> are what's giving you the trouble you mention in the comments. With them, this code:</p> <pre><code>`(define ClassName (lambda (args) (let (,(get-ivars '(iVar1 iVar2 iVar3) 1)) ;; your method-getting code ))) </code></pre> <p>Gives you this:</p> <pre><code>(define ClassName (lambda (args) (let (((iVar1 (nth args 1)) (iVar2 (nth args 2)) (iVar3 (nth args 3)))) ;; method-getting code ))) </code></pre> <p>Which, as you can see, gives you an extra set of parentheses around the assignments in the let.</p> <p>So you want to do this:</p> <pre><code>`(define ClassName (lambda (args) (let ,(get-ivars '(iVar1 iVar2 iVar3) 1) ;; your method-getting code ))) </code></pre> <p><code>get-ivars</code> is returning a list of lists, which is exactly what you want for the assignments in the <code>let</code>, so you don't need to wrap or (as I had it earlier) splice it. Just use the unquote on its own, and the result is:</p> <pre><code>(define ClassName (lambda (args) (let ((iVar1 (nth args 1)) (iVar2 (nth args 2)) (iVar3 (nth args 3))) ;; method-getting code ))) </code></pre> <p>Which should do the trick.</p> <p>Incidentally, I found it helpful to leave off the <code>eval</code> when I was playing around with this; one can then visually inspect the result to make sure its syntax is okay. </p>
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    1. POImproperly placed closed parenthesis in scheme function
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    1. USJosh Caswell
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    1. COI believe the `cons` here screws things up (take a look at my answer) -- he *does* want the `nil` at the end to stay, and he doesn't want things to get appended together. But I could be wrong too... I haven't tested either.
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    2. COWell, this returns what I want, but maybe I wasn't sure what I wanted. I'm passing the value to a (let ...), but the let spits out "bad syntax (not an identifier) in: (x (nth args 1))". The function you provided returns: ((x (nth args 1)) (y (nth args 2)))
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    3. CO@Mehrdad: Oh, I see. You're quite right. I hadn't registered that -- it seems strange to want to keep the `'()`. I've just tested both of our functions and they work, the only difference being the `'()` at the end.
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