> The result of the third instruction above is 14.5.
> According to the way I
> read the ANINT function the fifth instruction should
> give me an answer of 15
> since VDIFF is equal to 14.5.
In the help file it doesn't say what ANINT will do
with a .5 value, but trying your code sure looks like
it's truncing that value instead of rounding up (doing
it with 14.5001 does produce 15).
You can get the result you're looking for with
(INT(.vDiff + .5))
--
Larry
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