Return remainder after subtraction

[connexion]

Perhaps my brain is still awash with beer!?

I was sure that there was a simple function (a bit like datediff) that returns the remainder when one number is subtracted from another.

I thought i had this burried in a database somewhere but can't seem to find
it. Can anyone give me a hint as to what it's called please?

Many Thanks

Vince

 

[Brianwarnock]

I take it you meant division MOD

Brian

 

[raskew]

Hi -

Would the eval() function provide what you're after?
From the debug window:

x = 99
y = 34
? cstr(eval(x-y) * 2)
130

Bob

 

[allan57]

There is no built in function to return the remainder of a subtraction, below is a simple function that will give you the remainder.

Public Function MinusRemainderCalculation(FirstNumber As Double, SecondNumber As Double) As Double

On Error Resume Next

MinusRemainderCalculation = (FirstNumber - SecondNumber) - Fix(FirstNumber - SecondNumber)

End Function

Private Sub Form_Load()

Dim dblGetResult As Double

On Error Resume Next

dblGetResult = MinusCalculationRemainder(12.34, 6.22)

MsgBox dblGetResult 'Displays a result of 0.12

End Sub

 

[Sergeant]

Well, I'm confused. I thought the 'remainder' in subtraction was called the DIFFERENCE, and you get it like this:
7-5=2
7 minus 5 equals 2 (the difference).

I really don't see the correlation between 12.34, 6.22, and 0.12.

 

[allan57]

Well, I'm confused. I thought the 'remainder' in subtraction was called the DIFFERENCE, and you get it like this:
7-5=2
7 minus 5 equals 2 (the difference).

Hi Sergeant

Yes you are correct, but what I have posted is a substitute of the VB MOD function, but instead of it being used by division it used by subtraction. As this is how I understood the original question to be.

Allan

 

[Sergeant]

I understand the assumption that he is talking about a modulus. That still doesn't explain this:

I really don't see the correlation between 12.34, 6.22, and 0.12.

 

[allan57]

Hi Sergeant

Now you come to mention it neither do I, dont know what I was on when I wrote that one. I had it in my head to give him just the decimal side of the remainder. Sorrow to have caused you so much confusion .

Allan

 

[Brianwarnock]

I'm getting too old for this, I think I'll go and lie down in a dark room.

Brian

 

[david.brent]

I'm not really a maths person but here goes...

7 divided by 4 = 1.75

the remainder = 0.75

times this by 4 and you get 3.

12 divided by 5 = 2.4

the renainder = 0.4

times this by 5 and you get 2.

See were I'm going with this?

Dim Num1 as integer, Num2 as integer
Dim Whole as integer, Remain as integer

Whole = int(Num1/Num2)
Reamin = Num2 * ((Num1/Num2) - Whole)

Something like that anyway (can't check as Access needs reinstalling on my PC)

Take care.

 

[modest]

The original post was really vague and the terminology was either incorrect or lacking additional substance.

You don't need a function or anything special for the difference in subtraction, so I'll assume you mean the remainder. You can use Mod (it's the built-in version of what David.Brent was expessing):

Remainder = num1 Mod num2

From the help file:
The modulus, or remainder, operator divides number1 by number2 (rounding floating-point numbers to integers) and returns only the remainder as result