i=m.i*1.00 isn't integer so it'll be stored as float.


Well, you can illuminate to me where this conversion 4/8bytes is useful?

VFP puts the N(x, y) values within an IEEE value,
and this introduces very many problems of rounding in sums
(as an example 0.1 cannot be represented exactly with an IEEE value)
and it is taken care to exactly represent a I32 value that is perfectly represented
with the IEEE values.

Sorry Sergey, this programming is a little schizophrenic.

Fabio