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copied!<p>Finding all matches is very similar to finding a random match. Below is a simple modification of the logic that generates random matches on <a href="http://www.debuggex.com" rel="nofollow">www.debuggex.com</a>, assuming you already have a parse tree.</p> <p>The idea is that for every subtree, you return a list of all possible generated strings, given a string that was generated by all previous nodes in your parse tree.</p> <pre><code>AltTree.all = (prefix) -> rets = [] for child in children rets.extend(child.all(prefix)) ConcatTree.all = (prefix) -> prefixes = [prefix] for child in children newPrefixes = [] for p in prefixes newPrefixes.extend(child.all(p)) prefixes = newPrefixes return prefixes RepeatTree.all = (prefix) -> prefixes = [prefix] rets = [] for i up to max newPrefixes = [] for p in prefixes newPrefixes.extend(onlyChild.all(p)) prefixes = newPrefixes if i >= min rets.extend(prefixes) return rets CharsetTree.all = (prefix) -> rets = [] for char in allValidChars(): rets.push(prefix + char) return rets </code></pre> <p>The rest of the trees are left as exercises (most notably the literal tree). </p> <p>Note that there are intentionally no optimizations for the sake of clarity. Calling <code>myTree.all('')</code> will generate a list such that every valid matching string appears once <strong>for every path that generates this string</strong>. You will probably want to add deduplication and get rid of the excessive copying.</p> <p>I should also add that this will only work for regular expressions that have a <em>small</em> number of total matching strings. This is because all of the strings are being stored. If you want to get around this limitation, you can <code>yield</code>ify this algorithm. You will need to maintain a stack (think of it as being a bread crumb trail) of where you are in the tree. When a new string is asked for, you will create it from the path you travelled, and then update the path.</p>

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