Why Does \'test\'.replaceAll(\'.\', \'a\') Result in \'aa\' in Java?

String ReplaceAll Anomaly: Double Replacement with ".*"

In Java, the String.replaceAll() method performs a global search and replace operation on a given string using a regular expression. However, an unexpected behavior arises when using the ".*" regex pattern, leading to a double replacement issue.

The question poses a scenario where "test".replaceAll(".", "a") yields "aa" as the result. This is because . is a greedy quantifier that matches any number of characters (including zero), capturing the entire input string initially. As a result, the first replacement replaces the entire string with "a."

However, . can also match an empty string. After the initial replacement, the empty string remaining at the end of the input qualifies as a match for . as well. Thus, a second replacement occurs, replacing the empty string with "a," resulting in the final output "aa."

To prevent this double replacement issue, consider alternatives to .* such as . , which requires at least one character to match. Alternatively, using replaceFirst() will only replace the first occurrence, avoiding the situation where an empty string is matched a second time.

Interestingly, some regex engines do not exhibit this double replacement behavior. For example, GNU sed will consider the input exhausted after the first match, preventing further replacements. However, it is crucial to be aware of this potential anomaly when using ".*" with the String.replaceAll() method in Java.

The above is the detailed content of Why Does 'test'.replaceAll('.', 'a') Result in 'aa' in Java?. For more information, please follow other related articles on the PHP Chinese website!