Length of the String without using length() method

Question

How can I find the length of a String without using the length() method of String class?

Answer 1

• str.toCharArray().length should work.

• Or how about:

  str.lastIndexOf("")

  Probably even runs in constant time :)

• Another one

  Matcher m = Pattern.compile("$").matcher(str);
  m.find();
  int length = m.end();
  

• One of the dumbest solutions: str.split("").length - 1

• Is this cheating: new StringBuilder(str).length()? :-)

Answer 2

String blah = "HellO";
int count = 0;
for (char c : blah.toCharArray()) {
    count++;
}
System.out.println("blah's length: " + count);

Answer 3

Since nobody's posted the naughty back door way yet:

public int getLength(String arg) {
  Field count = String.class.getDeclaredField("count");
  count.setAccessible(true); //may throw security exception in "real" environment
  return count.getInt(arg);
}

;)

Answer 4

You can use a loop to check every character position and catch the IndexOutOfBoundsException when you pass the last character. But why?

public int slowLength(String myString) {
    int i = 0;
    try {
        while (true) {
            myString.charAt(i);
            i++;
        }
    } catch (IndexOutOfBoundsException e) {
       return i;
    }
}

Note: This is very bad programming practice and very inefficient.

You can use reflection to examine the internal variables in the String class, specifically count.

Answer 5

Just to complete this with the most stupid method I can come up with: Generate all possible strings of length 1, use equals to compare them to the original string; if they are equal, the string length is 1. If no string matches, generate all possible strings of length 2, compare them, for string length 2. Etc. Continue until you find the string length or the universe ends, whatever happens first.

Answer 6

try below code

    public static int Length(String str) {
    str = str + '\0';
    int count = 0;

    for (int i = 0; str.charAt(i) != '\0'; i++) {
        count++;
    }

    return count;
    }

Answer 7

For the semi-best methods have been posted and there's nothing better then String#length...

Redirect System.out to a FileOutputStream, use System.out.print (not println()!) to print the string and get the file size - this is equal to the string length. Don't forget to restore System.out after the measurement.

;-)

Answer 8

This is a complete program you can compile and run it.

import java.util.Scanner;

class Strlen{

    public static void main(String...args){
        Scanner sc = new Scanner(System.in);
        System.out.print("\nEnter Your Name =>" +"  ");
        String ab = sc.nextLine();
        System.out.println("\nName Length is:" +len(ab));
    }

    public static int len(String ab){
        char[] ac = ab.toCharArray();
        int i = 0, k = 0;

        try{
            for(i=0,k=0;ac[i]!='\0';i++)
                k++;
        }
        catch(Exception e){
        }
        return k;
    }

}

Answer 9

Hidden length() usage:

    String s = "foobar";

    int i = 0;
    for(char c: s.toCharArray())
    {
        i++;
    }

Answer 10

Here's another way:

int length = 0;
while (!str.equals("")) {
    str = str.substring(1);
    ++length;
}

In the same spirit (although much less efficient):

String regex = "(?s)";
int length = 0;
while (!str.matches(regex)) {
    regex += ".";
    ++length;
}

Or even:

int length = 0;
while (!str.matches("(?s).{" + length + "}")) {
    ++length;
}

Answer 11

We can iterate through the string like a character array, and count that way (A much more down to earth way of doing it):

String s = "foo"
char arr[]=s.toCharArray();
int len = 0;
for(char single : arr){
  len++;
}

Using the "foreach" version of the for loop

Answer 12

Just for completeness (and this is not at all recommended):

int length;
try
{
   length = str.getBytes("UTF-16BE").length / 2
}
catch (UnsupportedEncodingException e)
{
   throw new AssertionError("Cannot happen: UTF-16BE is always a supported encoding");
}

This works because a char is a UTF-16 code unit, and str.length() returns the number of such code units. Each UTF-16 code unit takes up 2 bytes, so we divide by 2. Additionally, there is no byte order mark written with UTF-16BE.

Answer 13

Even more slower one

public int slowerLength(String myString) {
String[] str = myString.split("");
int lol=0;
for(String s:str){
    lol++;
}
return (lol-1)
}

Or even slower,

public int slowerLength(String myString) {
String[] str = myString.split("");
int lol=0;
for(String s:str){
    lol += s.toCharArray().length;
}
return lol
}

Answer 14

Very nice solutions. Here are some more.

int length ( String s )
{
     int length = 0 ;
     // iterate through all possible code points
     for ( int i = INTEGER . MIN_VALUE ; i <= INTEGER . MAX_VALUE ; i ++ ) 
     {
           // count the number of i's in the string
          for ( int next = s . indexOf ( i , next ) + 1 ; next != -1 ; next = s . indexOf ( i , next ) + 1 )
          {
               length ++ ;
          }
     }
     return ( length ) ;
}

Here is a recursive version:

int length ( String s )
{
     int length = 0 ;
     search :
     for ( int i = Integer . MIN_VALUE ; i <= Integer . MAX_VALUE ; i ++ )
     {
          final int k = s . indexOf ( i ) ;
          if ( k != -1 )
          {
               length = length ( s . substring ( 0 , k ) ) + length ( s . substring ( k ) ) ;
               break search ;
          }
     }
     return ( length ) ;
}

And still more

int length ( String s )
{
     int length ;
     search ;
     for ( length = 0 ; true ; length ++ )
     {
          int [ ] codePoints = new int [ length ] ;
          for ( each possible value of codePoints from {MIN_VALUE,MIN_VALUE,...} to {MAX_VALUE,MAX_VALUE,...} )
          {
               if ( new String ( codePoints ) . equals ( s ) ) { break search ; }
          }
     }
}

How could I forget one that actually works in a reasonable time? (String#length is still preferred.)

int length ( String s )
{
     String t = s . replaceAll ( "." , "A" ) ;
     int length ;
     String r = "" ;
     search :
     for ( r = "" , length = 0 ; true ; r += "A" , length ++ )
          {
               if ( r . equals ( t ) )
               {
                    break search ;
               }
          }
     return ( length ) ;
}