Re: Newbie Question



"Oliver Wong" <owong@xxxxxxxxxxxxxx> wrote in message
news:S1kfg.223$I61.46@xxxxxxxxxxx

It takes 34 bits to store an unsigned 10 digit number, and 35 bits to
store a signed 10 digit number, so you're right that LONG INTEGER (32 bit
signed) won't be enough to store all values of the PIC 9(10). However, if
you move to floating point numbers (e.g. SINGLE or DOUBLE), you're opening
yourself up to problems with rounding, which may also yield big headaches.

What additional problems could there be ..considering this application was
already MOVEing a PIC 9(10) field to a PIC 9(1) field.... seems to me any
rounding problems introduced with floating point numbers are kind of
immaterial .....

(Besides, a DOUBLE has 16 decimal digits of precision).


MCM










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