Greedy quantifier on a repeated pattern

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Accepted Answer

With greedy .+, the engine consumes 'XbXcX' and then still needs a 'd' at the end. Since the string ends in 'd', the entire 'aXbXcXd' matches. With lazy .+?, you might expect the shortest possible match. But .+? still requires at least one character, and then it tries to stop as early as possible. The first 'd' it can find is still the last character in the string, so the match is still 'aXbXcXd'. Lazy matching only produces shorter matches when there are multiple valid endpoints earlier in the string. When the target character appears only once, greedy and lazy give the same result. Change 'd' to 'X' and you would see a difference: greedy would grab as far as possible, lazy would stop at the first 'X'.