Nicolas Alvarez wrote:
> The problem with doing #declare V=V+1/100; is that it might not reach 1.0000
> exactly at the end of the loop (because of floating point inaccuracies), so
> it might run one time too many.
> 

Right. If you MUST use floats, more robust is this sort of thing:

#local xmin = ....;
#local xmax = ....;
#local dx = .....;

#local x = xmin;
#while (x < xmax + dx / 2)
   #local x = x + dx;
#end

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