How can I normalize the EOL character in Java?

Question

I have a linux server and many clients with many operating systems. The server takes an input file from clients. Linux has end of line char LF, while Mac has end of line cha

Answer 1

Although String.replaceAll() is simpler to code, this should perform better since it doesn't go through the regex infrastructure.

    /**
 * Accepts a non-null string and returns the string with all end-of-lines
 * normalized to a \n.  This means \r\n and \r will both be normalized to \n.
 * 

 *     Impl Notes:  Although regex would have been easier to code, this approach
 *     will be more efficient since it's purpose built for this use case.  Note we only
 *     construct a new StringBuilder and start appending to it if there are new end-of-lines
 *     to be normalized found in the string.  If there are no end-of-lines to be replaced
 *     found in the string, this will simply return the input value.
 * 
 *
 * @param inputValue !null, input value that may or may not contain new lines
 * @return the input value that has new lines normalized
 */
static String normalizeNewLines(String inputValue){
    StringBuilder stringBuilder = null;
    int index = 0;
    int len = inputValue.length();
    while (index < len){
        char c = inputValue.charAt(index);
        if (c == '\r'){
            if (stringBuilder == null){
                stringBuilder = new StringBuilder();
                // build up the string builder so it contains all the prior characters
                stringBuilder.append(inputValue.substring(0, index));
            }
            if ((index + 1 < len) &&
                inputValue.charAt(index + 1) == '\n'){
                // this means we encountered a \r\n  ... move index forward one more character
                index++;
            }
            stringBuilder.append('\n');
        }else{
            if (stringBuilder != null){
                stringBuilder.append(c);
            }
        }
        index++;
    }
    return stringBuilder == null ? inputValue : stringBuilder.toString();
}