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HomeJavajavaTutorialWhy does `String.replaceAll(\'.*\', \'a\')` result in \'aa\' instead of \'a\'?

Why does `String.replaceAll(

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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