Mod with Doubles

As a work around your can do some simple math on the values. To get two decimal of precision just multiply the input values by 100 and then divide the result by 100.

result = (123.45*100 Mod 1*100)/100
result = (12345 Mod 100)/100
result = 0.45

I'm late to the party, but just incase this answer is still helpful to someone.

According to the VB6/VBA documentation

The modulus, or remainder, operator divides number1 by number2 (rounding floating-point numbers to integers) and returns only the remainder as result. For example, in the following expression, A (result) equals 5. A = 19 Mod 6.7 Usually, the data type of result is a Byte, Byte variant, Integer, Integer variant, Long, or Variant containing a Long, regardless of whether or not result is a whole number. Any fractional portion is truncated. However, if any expression is Null, result is Null. Any expression that is Empty is treated as 0.

Remember, mod returns the remainder of the division. Any integer mod 1 = 0.

debug.print 12 mod 1 
'12/1 = 12 r 0

The real culprit here though is that vba truncates (rounds down) the double to an integer before performing the modulo.

?13 mod 10
 '==>3 
?12.5 mod 10
 '==>2 

debug.print 12.5 mod 1
'vba truncates 12.5 to 12
debug.print 12 mod 1
'==> 0

I believe that the Mod operator calculates with long type only. The link that you provided is for VB.Net, which is not the same as the VBA you use in MSAccess.

The operator in VBA appears to accept a double type, but simply converts it to a long internally.

This test yielded a result of 1.

9 Mod 4.5

This test yielded a result of 0.

8 Mod 4.5