Length of the String without using length() method [closed]

Answer 1

Hidden length() usage:

    String s = "foobar";

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

Answer 2

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 3

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 ) ;
}

Answer 4

• 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 5

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 6

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