What is the easiest/best/most correct way to iterate through the characters of a string in Java?

Question

Some ways to iterate through the characters of a string in Java are:

1. Using StringTokenizer?
2. Converting the String to a char[] and iterating over that.

What is the easiest/best/most correct way to iterate?

Depreciable

See also #1528356 – Lipetsk

See also #8894758 Benchmarks show String.charAt() is fastest for small strings, and using reflection to read the char array directly is fastest for large strings. – Indianapolis

See also How do I turn a String into a Stream in java? – Carrnan

Java 8 : https://mcmap.net/q/79835/-what-is-the-easiest-best-most-correct-way-to-iterate-through-the-characters-of-a-string-in-java – Maxey

There are a countless ways to write, and implement, an algorithm for traversing a string, char by char, in Java. Which one is most correct, easist, and most simple are 3 different questions, and the answer for any of those 3 questions would be contingent on the programs environment, the data in the strings, and the reason for traversing the string. And even if you gave me all that information, any answer that I could give you, would be an opinion, it would be what I felt was the easiest most correct — "most what ever else you said" — way of doing it. – Layoff

This post has over 400+ votes, and none of those people ever thought once that this post should be flagged, thats sad... This question needs to be closed untill its edited, but that to is my opinion. – Layoff

Answer 1

I use a for loop to iterate the string and use charAt() to get each character to examine it. Since the String is implemented with an array, the charAt() method is a constant time operation.

String s = "...stuff...";

for (int i = 0; i < s.length(); i++){
    char c = s.charAt(i);        
    //Process char
}

That's what I would do. It seems the easiest to me.

As far as correctness goes, I don't believe that exists here. It is all based on your personal style.

Specific

Does the compiler inline the length() method? – Crimple

I dunno. I usually don't optimize my code. But it can't hurt to pull the length into a variable and use that instead. My guess is that the compiler in-lines the call though. – Specific

@Uri, the Java compiler does not do optimization. For HotSpot the JVM will inline it pretty soon at runtime. There are other JVM implementations (i.e. some of the J2ME VMs used in phones) that do not do runtime optimizations. – Orthogenesis

it might inline length(), that is hoist the method behind that call up a few frames, but its more efficient to do this for(int i = 0, n = s.length() ; i < n ; i++) { char c = s.charAt(i); } – Sapheaded

Cluttering your code for a tiny performance gain. Please avoid this until you decide this area of code is speed-critical. – Fireman

I usually don't optimize my code unless readability isn't sacrificed. – Specific

Note that this technique gives you characters, not code points, meaning you may get surrogates. – Blondell

charAt is not O(1) - it's O(N) for surrogates. – Overunder

@slim: Which clutter are you advising people to avoid -- caching the length using n? Or using an i loop instead of the for-each construct? – Hypersensitize

@larsH in this case I was talking about the n but I would also usually code a construct that didn't use the i either. – Fireman

@Overunder charAt is not O(1): How is that so? The code for String.charAt(int) is merely doing value[index]. I think you're confusing chatAt() with something else that give you code points. – Ichthyosis

what if length of String is more then the range of int? – Partner

@Indermalviya the maximum length of a String is Integer.MAX_VALUE – Bangka

Answer 2

Two options

for(int i = 0, n = s.length() ; i < n ; i++) { 
    char c = s.charAt(i); 
}

or

for(char c : s.toCharArray()) {
    // process c
}

The first is probably faster, then 2nd is probably more readable.

Sapheaded

plus one for placing the s.length() in the initialization expression. If anyone doesn't know why, it's because that is only evaluated once where if it was placed in the termination statement as i < s.length(), then s.length() would be called each time it looped. – Kuo

I thought compiler optimization took care of that for you. – Beyer

Any further thoughts on this? Can we reasonably expect compiler optimization to take care of avoiding the repeated call to s.length(), or not? – Glasser

@Glasser You can use the Javap class disassembler to see that the repeated calls to s.length() in for loop termination expression are indeed avoided. Note that in the code OP posted the call to s.length() is in the initialization expression, so the language semantics already guarantees that it will be called only once. – Schweiz

Also, see #197330 – Synthiasyntonic

@prasopes Note though that most java optimizations happen in the runtime, NOT in the class files. Even if you saw repeated calls to length() that doesn't indicate a runtime penalty, necessarily. – Viridi

@DaveCheney, why would you define 'n = s.length()' instead just have '(int i = 0; i<s.length(); i++){' ? – Condillac

@Lasse, the putative reason is for efficiency - your version calls the length() method on every iteration, whereas Dave calls it once in the initializer. That said, it is very likely the JIT ("just in time") optimizer will optimize the extra call away, so it's likely only a readability difference for no real gain. – Bechtold

And in my opinion, @Steve, it's actually less readable because (1) it's unconventional so it will distract people reading your code (as it did Lasse and many of the commenters), and (2) it moves the declaration away from its use. – Nahama

I dont get why would the first be faster. I thought foreach is the best optimized for performance solution, isn't it? – Voracious

toCharArray copies the contents of the String into a new array, which you avoid using charAt with a regular for loop. – Desrosiers

Also regarding the s.length() call in the initializer, that's also textbook premature optimization that hinders readability... isn't it? – Medium

Answer 3

Note most of the other techniques described here break down if you're dealing with characters outside of the BMP (Unicode Basic Multilingual Plane), i.e. code points that are outside of the u0000-uFFFF range. This will only happen rarely, since the code points outside this are mostly assigned to dead languages. But there are some useful characters outside this, for example some code points used for mathematical notation, and some used to encode proper names in Chinese.

In that case your code will be:

String str = "....";
int offset = 0, strLen = str.length();
while (offset < strLen) {
  int curChar = str.codePointAt(offset);
  offset += Character.charCount(curChar);
  // do something with curChar
}

The Character.charCount(int) method requires Java 5+.

Source: http://mindprod.com/jgloss/codepoint.html

Gyrus

I don't get how you use anything but the Basic Multilingual Plane here. curChar is still 16 bits righ? – Sedition

You either use an int to store the entire code point, or else each char will only store one out of the two surrogate pairs that define the code point. – Gyrus

I think I need to read up on code points and surrogate pairs. Thanks! – Sedition

+1 since this seems to be the only answer that is correct for Unicode chars outside of the BMP – Presume

Wrote some code to illustrate the concept of iterating over codepoints (as opposed to chars): gist.github.com/EmmanuelOga/… – Synthiasyntonic

Important point, and specifically asked at: #1528356 – Paving

Answer 4

In Java 8 we can solve it as:

String str = "xyz";
str.chars().forEachOrdered(i -> System.out.print((char)i));
str.codePoints().forEachOrdered(i -> System.out.print((char)i));

The method chars() returns an IntStream as mentioned in doc:

Returns a stream of int zero-extending the char values from this sequence. Any char which maps to a surrogate code point is passed through uninterpreted. If the sequence is mutated while the stream is being read, the result is undefined.

The method codePoints() also returns an IntStream as per doc:

Returns a stream of code point values from this sequence. Any surrogate pairs encountered in the sequence are combined as if by Character.toCodePoint and the result is passed to the stream. Any other code units, including ordinary BMP characters, unpaired surrogates, and undefined code units, are zero-extended to int values which are then passed to the stream.

How is char and code point different? As mentioned in this article:

Unicode 3.1 added supplementary characters, bringing the total number of characters to more than the 2^16 = 65536 characters that can be distinguished by a single 16-bit char. Therefore, a char value no longer has a one-to-one mapping to the fundamental semantic unit in Unicode. JDK 5 was updated to support the larger set of character values. Instead of changing the definition of the char type, some of the new supplementary characters are represented by a surrogate pair of two char values. To reduce naming confusion, a code point will be used to refer to the number that represents a particular Unicode character, including supplementary ones.

Finally why forEachOrdered and not forEach ?

The behaviour of forEach is explicitly nondeterministic where as the forEachOrdered performs an action for each element of this stream, in the encounter order of the stream if the stream has a defined encounter order. So forEach does not guarantee that the order would be kept. Also check this question for more.

For difference between a character, a code point, a glyph and a grapheme check this question.

Maxey

I think this is the most up-to-date answer here. – Howsoever

Answer 5

I agree that StringTokenizer is overkill here. Actually I tried out the suggestions above and took the time.

My test was fairly simple: create a StringBuilder with about a million characters, convert it to a String, and traverse each of them with charAt() / after converting to a char array / with a CharacterIterator a thousand times (of course making sure to do something on the string so the compiler can't optimize away the whole loop :-) ).

The result on my 2.6 GHz Powerbook (that's a mac :-) ) and JDK 1.5:

• Test 1: charAt + String --> 3138msec
• Test 2: String converted to array --> 9568msec
• Test 3: StringBuilder charAt --> 3536msec
• Test 4: CharacterIterator and String --> 12151msec

As the results are significantly different, the most straightforward way also seems to be the fastest one. Interestingly, charAt() of a StringBuilder seems to be slightly slower than the one of String.

BTW I suggest not to use CharacterIterator as I consider its abuse of the '\uFFFF' character as "end of iteration" a really awful hack. In big projects there's always two guys that use the same kind of hack for two different purposes and the code crashes really mysteriously.

Here's one of the tests:

    int count = 1000;
    ...

    System.out.println("Test 1: charAt + String");
    long t = System.currentTimeMillis();
    int sum=0;
    for (int i=0; i<count; i++) {
        int len = str.length();
        for (int j=0; j<len; j++) {
            if (str.charAt(j) == 'b')
                sum = sum + 1;
        }
    }
    t = System.currentTimeMillis()-t;
    System.out.println("result: "+ sum + " after " + t + "msec");

Recurrent

This has the same problem outlined here: #197330 – Synthiasyntonic

Answer 6

There are some dedicated classes for this:

import java.text.*;

final CharacterIterator it = new StringCharacterIterator(s);
for(char c = it.first(); c != CharacterIterator.DONE; c = it.next()) {
   // process c
   ...
}

Unscramble

Looks like an overkill for something as simple as iterating over immutable char array. – Orthogenesis

I don't see why this is overkill. Iterators are the most java-ish way to do anything... iterative. The StringCharacterIterator is bound to take full advantage of immutability. – Fireman

If I were using an iterator I would have used a foreach loop then. – Specific

@jjnguy: foreach is only possible for java.lang.Iterable's – Unscramble

Agree with @Orthogenesis - this is overkill. The only reason to use an iterator would be to take advantage of foreach, which is a bit easier to "see" than a for loop. If you're going to write a conventional for loop anyway, then might as well use charAt() – Choplogic

Using the character iterator is probably the only correct way to iterate over characters, because Unicode requires more space than a Java char provides. A Java char contains 16 bit and can hold Unicode characters up U+FFFF but Unicode specifies characters up to U+10FFFF. Using 16 bits to encode Unicode results in a variable length character encoding. Most answers on this page assume that the Java encoding is a constant length encoding, which is wrong. – Homologous

@Homologous It does not seem that a character iterator is going to help you with non-BMP characters: oracle.com/us/technologies/java/supplementary-142654.html – Unscramble

Answer 7

If you have Guava on your classpath, the following is a pretty readable alternative. Guava even has a fairly sensible custom List implementation for this case, so this shouldn't be inefficient.

for(char c : Lists.charactersOf(yourString)) {
    // Do whatever you want     
}

UPDATE: As @Alex noted, with Java 8 there's also CharSequence#chars to use. Even the type is IntStream, so it can be mapped to chars like:

yourString.chars()
        .mapToObj(c -> Character.valueOf((char) c))
        .forEach(c -> System.out.println(c)); // Or whatever you want

Thrower

If you need to do anything complex then go with the for loop + guava since you can't mutate variables (e.g. Integers and Strings) defined outside the scope of the forEach inside the forEach. Whatever is inside the forEach also can't throw checked exceptions, so that's sometimes annoying also. – Overmatch

Answer 8

If you need to iterate through the code points of a String (see this answer) a shorter / more readable way is to use the CharSequence#codePoints method added in Java 8:

for(int c : string.codePoints().toArray()){
    ...
}

or using the stream directly instead of a for loop:

string.codePoints().forEach(c -> ...);

There is also CharSequence#chars if you want a stream of the characters (although it is an IntStream, since there is no CharStream).

Jesse

Answer 9

If you need performance, then you must test on your environment. No other way.

Here example code:

int tmp = 0;
String s = new String(new byte[64*1024]);
{
    long st = System.nanoTime();
    for(int i = 0, n = s.length(); i < n; i++) {
        tmp += s.charAt(i);
    }
    st = System.nanoTime() - st;
    System.out.println("1 " + st);
}

{
    long st = System.nanoTime();
    char[] ch = s.toCharArray();
    for(int i = 0, n = ch.length; i < n; i++) {
        tmp += ch[i];
    }
    st = System.nanoTime() - st;
    System.out.println("2 " + st);
}
{
    long st = System.nanoTime();
    for(char c : s.toCharArray()) {
        tmp += c;
    }
    st = System.nanoTime() - st;
    System.out.println("3 " + st);
}
System.out.println("" + tmp);

On Java online I get:

1 10349420
2 526130
3 484200
0

On Android x86 API 17 I get:

1 9122107
2 13486911
3 12700778
0

Waylen

Benchmarks like these aren't reliable due to how the JVM works (e.g. optimizations and JIT). You would need to use JMH to get useful numbers here. – Ethnology

Answer 10

I wouldn't use StringTokenizer as it is one of classes in the JDK that's legacy.

The javadoc says:

StringTokenizer is a legacy class that is retained for compatibility reasons although its use is discouraged in new code. It is recommended that anyone seeking this functionality use the split method of String or the java.util.regex package instead.

Jaehne

String tokenizer is perfectly valid (and more efficient) way for iterating over tokens (i.e. words in a sentence.) It is definitely an overkill for iterating over chars. I am downvoting your comment as misleading. – Orthogenesis

ddimitrov: I'm not following how pointing out that StringTokenizer is not recommended INCLUDING a quotation from the JavaDoc (java.sun.com/javase/6/docs/api/java/util/StringTokenizer.html) for it stating as such is misleading. Upvoted to offset. – Coral

Thanks Mr. Bemrose ... I take it that the cited block quote should have been crystal clear, where one should probably infer that active bug fixes won't be commited to StringTokenizer. – Jaehne

Answer 11

public class Main {

public static void main(String[] args) {
     String myStr = "Hello";
     String myStr2 = "World";
      
     for (int i = 0; i < myStr.length(); i++) {    
            char result = myStr.charAt(i);
                 System.out.println(result);
     } 
        
     for (int i = 0; i < myStr2.length(); i++) {    
            char result = myStr2.charAt(i);
                 System.out.print(result);              
     }    
   }
}

Output:

H
e
l
l
o
World

Hairdo

Answer 12

See The Java Tutorials: Strings.

public class StringDemo {
    public static void main(String[] args) {
        String palindrome = "Dot saw I was Tod";
        int len = palindrome.length();
        char[] tempCharArray = new char[len];
        char[] charArray = new char[len];

        // put original string in an array of chars
        for (int i = 0; i < len; i++) {
            tempCharArray[i] = palindrome.charAt(i);
        } 

        // reverse array of chars
        for (int j = 0; j < len; j++) {
            charArray[j] = tempCharArray[len - 1 - j];
        }

        String reversePalindrome =  new String(charArray);
        System.out.println(reversePalindrome);
    }
}

Put the length into int len and use for loop.

Tyrosine

I'm starting to feel a bit spammerish... if there's such a word :). But this solution also has the problem outlined here: This has the same problem outlined here: #197330 – Synthiasyntonic

Answer 13

StringTokenizer is totally unsuited to the task of breaking a string into its individual characters. With String#split() you can do that easily by using a regex that matches nothing, e.g.:

String[] theChars = str.split("|");

But StringTokenizer doesn't use regexes, and there's no delimiter string you can specify that will match the nothing between characters. There is one cute little hack you can use to accomplish the same thing: use the string itself as the delimiter string (making every character in it a delimiter) and have it return the delimiters:

StringTokenizer st = new StringTokenizer(str, str, true);

However, I only mention these options for the purpose of dismissing them. Both techniques break the original string into one-character strings instead of char primitives, and both involve a great deal of overhead in the form of object creation and string manipulation. Compare that to calling charAt() in a for loop, which incurs virtually no overhead.

Napoleonnapoleonic

Answer 14

Elaborating on this answer and this answer.

Above answers point out the problem of many of the solutions here which don't iterate by code point value -- they would have trouble with any surrogate chars. The java docs also outline the issue here (see "Unicode Character Representations"). Anyhow, here's some code that uses some actual surrogate chars from the supplementary Unicode set, and converts them back to a String. Note that .toChars() returns an array of chars: if you're dealing with surrogates, you'll necessarily have two chars. This code should work for any Unicode character.

    String supplementary = "Some Supplementary: 𠜎𠜱𠝹𠱓";
    supplementary.codePoints().forEach(cp -> 
            System.out.print(new String(Character.toChars(cp))));

Castroprauxel

Answer 15

This Example Code will Help you out!

import java.util.Comparator;
import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;

public class Solution {
    public static void main(String[] args) {
        HashMap<String, Integer> map = new HashMap<String, Integer>();
        map.put("a", 10);
        map.put("b", 30);
        map.put("c", 50);
        map.put("d", 40);
        map.put("e", 20);
        System.out.println(map);

        Map sortedMap = sortByValue(map);
        System.out.println(sortedMap);
    }

    public static Map sortByValue(Map unsortedMap) {
        Map sortedMap = new TreeMap(new ValueComparator(unsortedMap));
        sortedMap.putAll(unsortedMap);
        return sortedMap;
    }

}

class ValueComparator implements Comparator {
    Map map;

    public ValueComparator(Map map) {
        this.map = map;
    }

    public int compare(Object keyA, Object keyB) {
        Comparable valueA = (Comparable) map.get(keyA);
        Comparable valueB = (Comparable) map.get(keyB);
        return valueB.compareTo(valueA);
    }
}

Limb

Answer 16

So typically there are two ways to iterate through string in java which has already been answered by multiple people here in this thread, just adding my version of it First is using

String s = sc.next() // assuming scanner class is defined above
for(int i=0; i<s.length(); i++){
     s.charAt(i)   // This being the first way and is a constant time operation will hardly add any overhead
  }

char[] str = new char[10];
str = s.toCharArray() // this is another way of doing so and it takes O(n) amount of time for copying contents from your string class to the character array

If performance is at stake then I will recommend using the first one in constant time, if it is not then going with the second one makes your work easier considering the immutability with string classes in java.

Mythological

Answer 17

If you need all characters one by one as String you can use this:

String text = "text";
for(String s: text.split("")) {

}

Froebel