25 lines
1003 B
Plaintext
25 lines
1003 B
Plaintext
When a back reference in a regex refers to a capturing group that hasn't been defined yet (or at all), it can never be matched. Named back references throw a ``++PatternSyntaxException++`` in that case; numeric back references fail silently when they can't match, simply making the match fail.
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When the group is defined before the back reference but on a different control path (like in ``++(.)|\1++`` for example), this also leads to a situation where the back reference can never match.
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== Noncompliant Code Example
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----
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Pattern.compile("\\1(.)"); // Noncompliant, group 1 is defined after the back reference
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Pattern.compile("(.)\\2"); // Noncompliant, group 2 isn't defined at all
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Pattern.compile("(.)|\\1"); // Noncompliant, group 1 and the back reference are in different branches
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Pattern.compile("(?<x>.)|\\k<x>"); // Noncompliant, group x and the back reference are in different branches
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----
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== Compliant Solution
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----
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Pattern.compile("(.)\\1");
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Pattern.compile("(?<x>.)\\k<x>");
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----
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