Re: Subsequent storage

new topic     » topic index » view thread      » older message » newer message 

Robert McDougal writes:
> Robert suggested using a temporary variable with this
> problem. So if I set t=x[i] append something to t, and set x[i]=t
> then that would take less time?  I'm more confused now about this
> then ever.  If so, then I don't understand why that is.  Wouldn't
> you still be dealing with the N-squared thing when you set x[i]=t?
> More confused now then ever,

I meant that something like:
     t = x[i]
     for i = 1 to 10000 do
          t = append(t, v)
     end for
     x[i] = t

will be much faster than appending to x[i] inside the loop.

Regards,
     Rob Craig
     Rapid Deployment Software
     http://members.aol.com/FilesEu/

new topic     » topic index » view thread      » older message » newer message

Search



Quick Links

User menu

Not signed in.

Misc Menu