Lets see what we can do to transform this:

for(x=x1;x <= x2;x++)
{
  <take an a step>
  if((eps << 1) >= DeltaX)
  {
    <take an b step>
    eps -= DeltaX;
  }
}

into something more effecient.

First

if ((eps << 1) >= DeltaX

Is (roughly, except for possible rounding issues) equivalent to:

if (eps >= DeltaX / 2)

But if we subtract DeltaX/2 from X before entering the loop then we have:

if (eps >= 0)

So now we have:

eps -= DeltaX >> 1;

for(x=x1;x <= x2;x++)
{
  <take an a step>
  if(eps >= 0)
  {
    <take an b step>
    eps -= DeltaX;
  }
}

Is that a little simpler?

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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