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copied!<p>We could use <strong>equal?</strong></p> <pre><code> (equal? '(a (b (c))) '(a (b (c)))) </code></pre> <p>But, for some fun, following on from Vassermans mention of a "break", this might be a good chance to take advantage of Schemes continuation controlling power!</p> <p>We can use <strong>call/cc</strong> to issue an early return if we notice any difference in the trees. This way we can just jump back to the callers continuation without having to unwind the stack.</p> <p>Here is a really simple example. It assumes the trees are well-formed and only contain symbols as leaves, but it should hopefully demonstrate the concept. You'll see that the procedure explicitly accepts the continuation as a parameter.</p> <pre><code> (define (same? a b return) (cond ((and (symbol? a) (symbol? b)) ; Both Symbols. Make sure they are the same. (if (not (eq? a b)) (return #f))) ((and (empty? a) (empty? b))) ; Both are empty, so far so good. ((not (eq? (empty? a) (empty? b))) ; One tree is empty, must be different! (return #f)) (else (begin (same? (car a) (car b) return) ; Lets keep on looking. (same? (cdr a) (cdr b) return))))) </code></pre> <p><strong>call/cc</strong> lets us capture the current continuation. Here is how I called this procedure:</p> <pre><code> (call/cc (lambda (k) (same? '(a (b)) '(a (b)) k))) ; --> #t (call/cc (lambda (k) (same? '(a (b (c) (d e))) '(a (b (c) (d e))) k))) ; --> #t (call/cc (lambda (k) (same? '(a (b (F) (d e))) '(a (b (c) (d e))) k))) ; --> #f (call/cc (lambda (k) (same? '(a (b)) '(a (b (c) (d))) k))) ; --> #f </code></pre>

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