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KL
 
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Harlan,

Yours is clearly a much better (and neater) implementation of the idea. As
to =32+1/3, I guess it is going to be the same issue for all possible
solutions given the IEEE 754 specification, so as long as one is aware of
that, =LEN(x)-LEN(INT(x))-1 is probably the best option.

Thanks and regards,
KL


"Harlan Grove" wrote in message
...
"Bob Phillips" wrote...
This doesn't work for me at all, it returns 15 for 34.45.

Looking at it MOD(ABS(A1),1) evaluated to .450000000000003, which I
would take is caused by lack of precision when using MOD.

...

The motivation may have been sound. The implementation wasn't. It should
be as simple as

=LEN(x)-LEN(INT(x))-1

though that'd work with values stored and used but not displayed.

What should the result be for, say, =32+1/3?