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Set flag/boolean when loop finish?

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  Set flag/boolean when loop finish? jonas.thornvall@gmail.com - 2014-07-17 10:29 -0700
    Re: Set flag/boolean when loop finish? jonas.thornvall@gmail.com - 2014-07-17 10:39 -0700
      Re: Set flag/boolean when loop finish? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2014-07-17 19:06 +0100
        Re: Set flag/boolean when loop finish? jonas.thornvall@gmail.com - 2014-07-17 11:18 -0700
        Re: Set flag/boolean when loop finish? jonas.thornvall@gmail.com - 2014-07-17 11:28 -0700
          Re: Set flag/boolean when loop finish? jonas.thornvall@gmail.com - 2014-07-17 11:40 -0700
            Re: Set flag/boolean when loop finish? jonas.thornvall@gmail.com - 2014-07-17 11:45 -0700

#25418 — Set flag/boolean when loop finish?

This code seem unecessarily set the flag to one multiple times, how can i condition the while to just set it as it step out of loop?
  
while (i * basemultip <= number && i < bas)
      {
         j = i;
         i ++ ;
         set = 1;
      }

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#25419

Den torsdagen den 17:e juli 2014 kl. 19:29:23 UTC+2 skrev jonas.t...@gmail.com:
> This code seem unecessarily set the flag to one multiple times, how can i condition the while to just set it as it step out of loop?
> 
>   
> 
> while (i * basemultip <= number && i < bas)
> 
>       {
> 
>          j = i;
> 
>          i ++ ;
> 
>          set = 1;
> 
>       }

Managed to reduce some conditional checks, well they was unnecessary to start with.

function enctwo(number, bas)
{
   basestr = "";
   basemultip = 1;
   digit = 1;

   while (basemultip < number)
   {
      basemultip = basemultip * bas;
      digit = digit + 1;
   }

   while (number > 0 && digit > 0)
   {
      set = 0;
      i = 0;
      while (i * basemultip <= number && i < bas)
      {
         j = i;
         i ++ ;
         set = 1;
      }
      if (set == 1)
      {
         subtrahend = basemultip * j;
         number = number - subtrahend;
         basestr = basestr + j + " ";
      }
      //else if (set == 0 && number != 0)basestr = basestr + 0 + " ";
      else {basestr = basestr + 0 + " ";}
      basemultip = basemultip / bas;
      digit -- ;
   }
   while (digit > 0)
   {
      basestr = basestr + 0 + " ";
      digit -- ;
   }
   return basestr;
}

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#25420

jonas.thornvall@gmail.com writes:
<snip>
> Managed to reduce some conditional checks, well they was unnecessary
> to start with.
<snip>
>       set = 0;
>       i = 0;
>       while (i * basemultip <= number && i < bas)
>       {
>          j = i;
>          i ++ ;
>          set = 1;
>       }
>       if (set == 1)

Here, the conditions set == 1 and i > 0 are the same.  There's no need
for 'set'.

<snip>
-- 
Ben.

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#25421

Den torsdagen den 17:e juli 2014 kl. 20:06:23 UTC+2 skrev Ben Bacarisse:
> jonas.thornvall@gmail.com writes:
> 
> <snip>
> 
> > Managed to reduce some conditional checks, well they was unnecessary
> 
> > to start with.
> 
> <snip>
> 
> >       set = 0;
> 
> >       i = 0;
> 
> >       while (i * basemultip <= number && i < bas)
> 
> >       {
> 
> >          j = i;
> 
> >          i ++ ;
> 
> >          set = 1;
> 
> >       }
> 
> >       if (set == 1)
> 
> 
> 
> Here, the conditions set == 1 and i > 0 are the same.  There's no need
> 
> for 'set'.
> 
> 
> 
> <snip>
> 
> -- 
> 
> Ben.

Of course, thanks Ben.

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#25422

Den torsdagen den 17:e juli 2014 kl. 20:06:23 UTC+2 skrev Ben Bacarisse:
> jonas.thornvall@gmail.com writes:
> 
> <snip>
> 
> > Managed to reduce some conditional checks, well they was unnecessary
> 
> > to start with.
> 
> <snip>
> 
> >       set = 0;
> 
> >       i = 0;
> 
> >       while (i * basemultip <= number && i < bas)
> 
> >       {
> 
> >          j = i;
> 
> >          i ++ ;
> 
> >          set = 1;
> 
> >       }
> 
> >       if (set == 1)
> 
> 
> 
> Here, the conditions set == 1 and i > 0 are the same.  There's no need
> 
> for 'set'.
> 
> 
> 
> <snip>
> 
> -- 
> 
> Ben.

Would it be possible to go direct to bijective code from this by reduce the previous digit when a zero appear and make next digit base, and if another 0 follow reduce that base value-1, sort of roll it off?

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#25423

Den torsdagen den 17:e juli 2014 kl. 20:28:50 UTC+2 skrev jonas.t...@gmail.com:
> Den torsdagen den 17:e juli 2014 kl. 20:06:23 UTC+2 skrev Ben Bacarisse:
> 
> > jonas.thornvall@gmail.com writes:
> 
> > 
> 
> > <snip>
> 
> > 
> 
> > > Managed to reduce some conditional checks, well they was unnecessary
> 
> > 
> 
> > > to start with.
> 
> > 
> 
> > <snip>
> 
> > 
> 
> > >       set = 0;
> 
> > 
> 
> > >       i = 0;
> 
> > 
> 
> > >       while (i * basemultip <= number && i < bas)
> 
> > 
> 
> > >       {
> 
> > 
> 
> > >          j = i;
> 
> > 
> 
> > >          i ++ ;
> 
> > 
> 
> > >          set = 1;
> 
> > 
> 
> > >       }
> 
> > 
> 
> > >       if (set == 1)
> 
> > 
> 
> > 
> 
> > 
> 
> > Here, the conditions set == 1 and i > 0 are the same.  There's no need
> 
> > 
> 
> > for 'set'.
> 
> > 
> 
> > 
> 
> > 
> 
> > <snip>
> 
> > 
> 
> > -- 
> 
> > 
> 
> > Ben.
> 
> 
> 
> Would it be possible to go direct to bijective code from this by reduce the previous digit when a zero appear and make next digit base, and if another 0 follow reduce that base value-1, sort of roll it off?

The nasty case would be when the leading number of the zeros is 1 i guess.

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#25424

Den torsdagen den 17:e juli 2014 kl. 20:40:36 UTC+2 skrev jonas.t...@gmail.com:
> Den torsdagen den 17:e juli 2014 kl. 20:28:50 UTC+2 skrev jonas.t...@gmail.com:
> 
> > Den torsdagen den 17:e juli 2014 kl. 20:06:23 UTC+2 skrev Ben Bacarisse:
> 
> > 
> 
> > > jonas.thornvall@gmail.com writes:
> 
> > 
> 
> > > 
> 
> > 
> 
> > > <snip>
> 
> > 
> 
> > > 
> 
> > 
> 
> > > > Managed to reduce some conditional checks, well they was unnecessary
> 
> > 
> 
> > > 
> 
> > 
> 
> > > > to start with.
> 
> > 
> 
> > > 
> 
> > 
> 
> > > <snip>
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >       set = 0;
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >       i = 0;
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >       while (i * basemultip <= number && i < bas)
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >       {
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >          j = i;
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >          i ++ ;
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >          set = 1;
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >       }
> 
> > 
> 
> > > 
> 
> > 
> 
> > > >       if (set == 1)
> 
> > 
> 
> > > 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > Here, the conditions set == 1 and i > 0 are the same.  There's no need
> 
> > 
> 
> > > 
> 
> > 
> 
> > > for 'set'.
> 
> > 
> 
> > > 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > <snip>
> 
> > 
> 
> > > 
> 
> > 
> 
> > > -- 
> 
> > 
> 
> > > 
> 
> > 
> 
> > > Ben.
> 
> > 
> 
> > 
> 
> > 
> 
> > Would it be possible to go direct to bijective code from this by reduce the previous digit when a zero appear and make next digit base, and if another 0 follow reduce that base value-1, sort of roll it off?
> 
> 
> 
> The nasty case would be when the leading number of the zeros is 1 i guess.

And an array approach rather then a string would be necessary.

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