> > I have a string of integer values:

> > p1<p2<p3<p4...<p8 and a reference value lndx

> > I have a code snippet that I have to do this way

> > begin p8 lndx < while
> >    0 p1 sieve c!  0 p2 sieve c!  0 p3 sieve c!  0 p4 sieve c!
> >    0 p5 sieve c!  0 p6 sieve c!  0 p7 sieve c!  0 p8 sieve c!
> >    x1 p1 + to p1   x1 p2 + to p2    x1 p3 + to p3    x1 p4 + to p4
> >    x1 p5 + to p5   x1 p6 + to p6   x1 p7 + to p7    x1 p8 + to p8
> > repeat
> > p7 lndx < if 0 p7 sieve c! then
> > p6 lndx < if 0 p6 sieve c! then
> > p5 lndx < if 0 p5 sieve c! then
> > p4 lndx < if 0 p4 sieve c! then
> > p3 lndx < if 0 p3 sieve c! then
> > p2 lndx < if 0 p2 sieve c! then
> > p1 lndx < if 0 p1 sieve c! then

> > So when p8 >= lndx I want to check the other pi < lndx once
> > more and if true, then do  0 pi sieve c! FOR ALL OTHER pi.

> > Conceptually, I want to do something like this:

> > p7 < lndx IF goto 7 elsif p6 < lndx if goto 6 elsif ...p1 < lndx goto
> > 1

> > 7: 0 p7 sieve c!
> > 6: 0 p6 sieve c!
> > 5: 0 p5 sieve c!
> > 4: 0 p4 sieve c!
> > 3: 0 p3 sieve c!
> > 2: 0 p2 sieve c!
> > 1: 0 p1 sieve c!

> > So, if pi < lndx I want to "goto" the ith label and do all the
> > operations
> > starting there to the end thru p1, since we know that if pi < lndx
> > then
> > all the pi-1,2,..etc < lndx will also be true.

> > The way I've done it was the shortest/fastest way I was able to do
> > this
> > correctly, writing, I believe, the smallest amount of repetitive code.

> > Is there a shorter/faster way to do this?

> Instead of p1 ... p8, use array p.

> Something like this (untested).

> Ruby:

> p = Array.new(8){ 0 }

> top = nil
> 7.downto(0){|i|
>   if p[i] < lndx
>     top = i
>     break
>   end}

> top.downto(0){|i|
>   sieve[ p[i] ] = 0

> }