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    <p>Besides what Hans Kesting already said, a lazy multiplier will do the exact oposite of the normal greedy multipliers: The possible match is kept as small as possible and the rest of the regular expression is tested.</p> <p>So if you’re having the string <code>aaba</code> and test the regular expression <code>a.*b</code> on it, the internal processing steps would be as follows:</p> <ol> <li><strong><code>a</code></strong> in <strong><code>a</code></strong><code>.*b</code> matches <strong><code>a</code></strong><code>aba</code></li> <li><strong><code>.*</code></strong> in <code>a</code><strong><code>.*</code></strong><code>b</code> matches <code>a</code><strong><code>a</code></strong><code>ba</code>, and since <code>.*</code> is greedy <ol start="3"> <li><strong><code>.*</code></strong> then matches <code>a</code><strong><code>ab</code></strong><code>a</code></li> <li><strong><code>.*</code></strong> then matches <code>a</code><strong><code>aba</code></strong></li> </ol></li> <li><strong><code>b</code></strong> in <code>a.*</code><strong><code>b</code></strong> fails as there is no letter left <ul> <li>backtracking goes one step back and <code>.*</code> will now only match <code>bb</code> in <code>a</code><strong><code>ab</code></strong><code>a</code></li> </ul></li> <li><strong><code>b</code></strong> in <code>a.*</code><strong><code>b</code></strong> still fails on <code>aab</code><strong><code>a</code></strong> <ul> <li>backtracking goes one step back and <code>.*</code> now matches only <code>b</code> in <code>a</code><strong><code>a</code></strong><code>ba</code></li> </ul></li> <li><strong><code>b</code></strong> in <code>a.*</code><strong><code>b</code></strong> now matches <code>b</code> in <code>aa</code><strong><code>b</code></strong><code>a</code> and we’re done.</li> </ol> <p>So the full match is <strong><code>aab</code></strong><code>a</code>.</p> <p>If we do the same with a lazy multiplier (<code>a.*?b</code>), the processing will do the oposite, try to match the least possible characters as possible:</p> <ol> <li><strong><code>a</code></strong> in <strong><code>a</code></strong><code>.*?b</code> matches <strong><code>a</code></strong><code>aba</code></li> <li><strong><code>.*</code></strong> in <code>a</code><strong><code>.*</code></strong><code>?b</code> matches nothing (<code>*</code> = zero or more repetitions), and since <code>.*</code> is declared as lazy (<code>.*?</code>), the rest of the regular expression is tested</li> <li><strong><code>b</code></strong> in <code>a.*?</code><strong><code>b</code></strong> fails on <code>a</code><strong><code>a</code></strong><code>ba</code> <ul> <li>backtracking will try to increase the match of <code>.*</code></li> </ul></li> <li><strong><code>.*</code></strong> matches now <code>a</code><strong><code>a</code></strong><code>ba</code></li> <li><strong><code>b</code></strong> in <code>a.*?</code><strong><code>b</code></strong> matches <code>aa</code><strong><code>b</code></strong><code>a</code> and we’re done.</li> </ol> <p>So the full match if <strong><code>aab</code></strong><code>a</code>.</p>
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    1. POHows does the Regular Expression: /.+?/ work?
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    1. POHows does the Regular Expression: /.+?/ work?
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    1. COSo a.*b would catch BOTH aab and ab, correct? But a.*?b only catches aab? (using your aaba example for both scenarios). So .*? isn't necessarily trying to find the LEAST possibly characters, but the first instance of '.' being 1 letter, and if it matches, done. So once a.*?b catches aab, it wouldn't go on to try catching just ab.
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    2. CO@hatorade: No, sorry. Both expressions match only **`aab`**`a`.
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    3. CO@Gumbo: I don't see why .*? wouldn't catch ab. a.*?b clearly matches ab. it makes sense that .* catches aab because it is greedy and tries to get the max fit. but isn't that the point of lazy plus?
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