Why does `String.replaceAll(\'.*\', \'a\')` result in \'aa\' instead of \'a\'?
Understanding String.replaceAll() Double Replacement Anomaly
The question arises as to why the following code:
<code class="java">System.out.println("test".replaceAll(".*", "a"));</code>
results in "aa" instead of the expected "a". This anomaly also occurs when using ".*$".
Cause:
The anomaly arises from the nature of the ".*" regex. It matches any character sequence, including an empty string. Therefore:
- First Match: “.*” matches the entire string "test" and replaces it with "a."
- Second Match: “.*” can also match an empty string, which it finds at the end of the input. This second match replaces the empty string with "a," resulting in the double replacement.
Solution:
To avoid this issue, consider using:
- .replaceFirst(): Only replaces the first occurrence, preventing the double replacement.
- . : Matches any character sequence of one or more characters, excluding empty strings.
Regex Behavior:
While .* can match an empty string, it cannot match more than twice. This is because:
- After the first match, the regex engine shifts one character ahead.
- The second match uses the .* to match the empty string that is now at the end of the input.
- The regex engine exhausts the input and shifts another character ahead.
- There is no more input for .* to match, preventing a third replacement.
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