Does Java's toLowerCase() preserve original string length?

Question

Assume two Java String objects:

String str = "<my string>"; String strLower = str.toLowerCase(); 

Is it then true that for every value of <my string> the expression

str.length() == strLower.length() 

evaluates to true?

So, does String.toLowerCase() preserve original string length for any value of String?

Answer 1

Surprisingly it does not!!

From Java docs of toLowerCase

Converts all of the characters in this String to lower case using the rules of the given Locale. Case mapping is based on the Unicode Standard version specified by the Character class. Since case mappings are not always 1:1 char mappings, the resulting String may be a different length than the original String.

Example:

package com.stackoverflow.q2357315;  import java.util.Locale;  public class Test {     public static void main(String[] args) throws Exception {         Locale.setDefault(new Locale("lt"));         String s = "\u00cc";         System.out.println(s + " (" + s.length() + ")"); // Ì (1)         s = s.toLowerCase();         System.out.println(s + " (" + s.length() + ")"); // i̇̀ (3)     } } 

Answer 2

First of all, I'd like to point out that I absolutely agree with the (currently highest-rated) answer of @codaddict.

But I wanted to do an experiment, so here it is:

It's not a formal proof, but this code ran for me without ever reaching the inside of the if (using JDK 1.6.0 Update 16 on Ubuntu):

Edit: Here's some updated code that handles Locales as well:

import java.util.Locale;  public class ToLowerTester {     public final Locale locale;      public ToLowerTester(final Locale locale) {         this.locale = locale;     }      public String findFirstStrangeTwoLetterCombination() {         char[] b = new char[2];         for (char c1 = 0; c1 < Character.MAX_VALUE; c1++) {             b[0] = c1;             for (char c2 = 0; c2 < Character.MAX_VALUE; c2++) {                 b[1] = c2;                 final String string = new String(b);                 String lower = string.toLowerCase(locale);                 if (string.length() != lower.length()) {                     return string;                 }             }         }         return null;     }     public static void main(final String[] args) {         Locale[] locales;         if (args.length != 0) {             locales = new Locale[args.length];             for (int i=0; i<args.length; i++) {                 locales[i] = new Locale(args[i]);             }         } else {             locales = Locale.getAvailableLocales();         }         for (Locale locale : locales) {             System.out.println("Testing " + locale + "...");             String result = new ToLowerTester(locale).findFirstStrangeTwoLetterCombination();             if (result != null) {                 String lower = result.toLowerCase(locale);                 System.out.println("Found strange two letter combination for locale "                     + locale + ": <" + result + "> (" + result.length() + ") -> <"                     + lower + "> (" + lower.length() + ")");             }         }     } } 

Running that code with the locale names mentioned in the accepted answer will print some examples. Running it without an argument will try all available locales (and take quite a while!).

It's not extensive, because theoretically there could be multi-character Strings that behave differently, but it's a good first approximation.

Also note that many of the two-character combinations produced this way are probably invalid UTF-16, so the fact that nothing explodes in this code can only be blamed on a very robust String API in Java.

And last but not least: even if the assumption is true for the current implementation of Java, that can easily change once future versions of Java implement future versions of the Unicode standard, in which the rules for new characters may introduce situations where this no longer holds true.

So depending on this is still a pretty bad idea.