View Single Post
  #7   Report Post  
KL
 
Posts: n/a
Default

Opps! You're right Bob. In my testing I hadn't run into this issue and I
didn't suspect any precision issue with MOD - it definetely returns
..450000000000003, which I believe has to do with the floating-point
limitations ( http://support.microsoft.com/kb/78113/en-us )

Regards,
KL


"Bob Phillips" wrote in message
...
KL,

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.

--

HTH

RP
(remove nothere from the email address if mailing direct)


"KL" wrote in message
...
...just in case the file might be used in other language environments,
the
following formula avoids using the decimal separator explicitly:

=IF(ISNUMBER(A1),LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0),0)

It also uses IF(ISNUMBER(A1),...,0) construct to check if there is text
in
the cell A1. If there can only be numbers or empty cells then a shorter
version can be used:

=LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0)

Regards,
KL



"Bob Phillips" wrote in message
...
=IF(ISNUMBER(FIND(".",A20)),LEN(A20)-FIND(".",A20),0)

--

HTH

RP
(remove nothere from the email address if mailing direct)


"Warren Smith" wrote in message
...
The worksheet I am doing needs to look at cell values, and in another
return
the number of decimal places used in the cell.

for example
if A1 contained a number of 34.45,
I want B1 to tell me 2.
I need to know this because I want to enter the data in a table
against
another set of figures which will have 1 more decimal place than the
first
set,
and I can't set the decimal places up before because I don't know how
many
there will be.

Thanks in advance for any help
Warren