Re: infinite recursion

From:
Roedy Green <see_website@mindprod.com.invalid>
Newsgroups:
comp.lang.java.help
Date:
Fri, 12 Oct 2007 06:38:29 GMT
Message-ID:
<hf5ug3h84fefen721phaa2go152u9qerpn@4ax.com>
On Thu, 11 Oct 2007 10:30:23 -0700, andreyvul <andrey.vul@gmail.com>
wrote, quoted or indirectly quoted someone who said :

short[] allOptions(short val) {
    Vector<Short> cc = new Vector<Short>(0);
    Iterator<Short> it;
    short[] rv;
    //check which numbers are available
    short i;
    for (i = 0; i < 9; ++i)
        //bitmask check (1 == unavailable, 0 == available)
        if (((1 << i) & val) == 0)
            //available, add number to cc
            cc.add(cc.size(), (short)(i + 1));
    //create return array
    rv = new short[cc.size()];
    it = cc.iterator();
    i = 0;
    while (it.hasNext())
        rv[i++] = it.next();
    return rv;
}


Here is your code tidied up a bit:
Vector -> ArrayList
use of for:each loop
one parm add
standard idiom for i
don't reuse i.

import java.util.ArrayList;
public class Temp
   {
   short[] allOptions( short val )
      {
      ArrayList<Short> cc = new ArrayList<Short>(10);

      // Check which numbers are available
      for ( int i = 0; i < 9; i++ )
         {
         // Bitmask check (1 == unavailable, 0 == available)
         if ( ( ( 1 << i ) & val ) == 0 )
            {
            // Available, add number to tail end of cc
            cc.add( (short)( i + 1 ) );
            }
         }
      // Create return array using for:each loop
      // We can't use toArray since it
      // will not accept short[] as a parm
      short[] rv = new short[ cc.size() ];
      int i = 0;
      for ( short s: cc )
         {
         rv[i++] = s;
         }
      return rv;
      }
   }

--
Roedy Green Canadian Mind Products
The Java Glossary
http://mindprod.com

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