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Why does `String.replaceAll(regex)` perform multiple replacements when the regex matches everything?

Linda Hamilton
Release: 2024-11-01 00:35:02
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Why does `String.replaceAll(regex)` perform multiple replacements when the regex matches everything?

Why String.replaceAll(regex) Performs Multiple Replacements for Regular Expressions Matching Everything

In Java, the String.replaceAll() method allows for the replacement of all occurrences of a regular expression pattern with a specified substitution. However, when the regular expression matches the entire input string, an unexpected behavior can occur, resulting in multiple replacements.

Consider the example presented:

<code class="java">System.out.println("test".replaceAll(".*", "a"));</code>
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This code outputs "aa" instead of the expected "a". This anomaly arises because the regular expression .* matches any string, including an empty string.

After the initial match, the regex engine attempts to locate another occurrence by starting from the end of the input. However, .* can match an empty string, and this match occurs at the very end of the input. Therefore, the regex engine substitutes an "a" at this empty position.

To avoid this behavior and ensure a single replacement, consider using the .* expression instead, which requires the regular expression to match at least one character and thus cannot match an empty string.

Alternatively, you can employ the String.replaceFirst() method, which only replaces the first occurrence of the specified regular expression.

For instance:

<code class="java">"test".replaceFirst(".*", "a")</code>
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This code will produce the desired output of "a".

It's worth mentioning that not all regex engines behave in the same manner regarding multiple matches. GNU sed, for example, will consider the input exhausted after the first match, preventing any further replacements.

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