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copied!<p>At macro expansion time, <code>cdr</code> is not evaluated. So <code>(count-true t t nil)</code> hits an infinite expansion like this:</p> <pre><code>(count-true t t nil) => (1+ (count-true (cdr (t t t nil)))) => (1+ (1+ (count-true (cdr (cdr (t t t nil)))))) => (1+ (1+ (1+ (count-true (cdr (cdr (cdr (t t t nil)))))))) => ... </code></pre> <p>Well, actually this happens for both recursive branches at once. So it blows up even faster than the example.</p> <p>A better idea? </p> <ol> <li><p>Trying writing the same thing as a function first. Figure out where you have to put the lambdas to delay evaluation. Then abstract the function into a macro so you can leave out the lambdas.</p> <p>The point of writing a function first, by the way, is that sometimes you'll find that a function is good enough. Writing a macro where a function would do is a mistake.</p></li> <li><p>Generally, when you write macros in Common Lisp, start with a <code>loop</code> rather than recursion. Recursive macros are tricky (and usually wrong :). </p></li> </ol> <p>Edit:</p> <p>Here is a more correct (but much longer) example:</p> <pre><code>(count-true t nil) => (cond ((null '(t nil)) 0) ((car '(t nil)) (1+ (count-true (cdr '(t nil))))) (T (count-true (cdr '(t nil))))) => (cond ((null '(t nil)) 0) ((car '(t nil)) (1+ (1+ (count-true (cdr (cdr '(t nil))))))) (T (count-true (cdr (cdr '(t nil)))))) => (cond ((null '(t nil)) 0) ((car '(t nil)) (1+ (1+ (1+ (count-true (cdr (cdr (cdr '(t nil))))))))) (T (count-true (cdr (cdr (cdr '(t nil))))))) => (cond ((null '(t nil)) 0) ((car '(t nil)) (1+ (1+ (1+ (1+ (count-true (cdr (cdr (cdr (cdr '(t nil))))))))))) (T (count-true (cdr (cdr (cdr (cdr '(t nil)))))))) </code></pre>

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