converting a negative number to a positive

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Hi
I want to convert a negative number to a positive.
the match.abs(dif) does not work.
Is there any other possible way to do this?
When i run the code as is ,if the dif is negative,it stays negative.

VB

Dim Total As Decimal = 0
               Dim currTotal As Decimal = Math.Round(dtreader(1), 2)
               Total = Math.Round(Decimal.Parse((dtreader(2) + dtreader(3)) + dtreader(4)), 2)
               Dim avg As Decimal = Math.Round(Total / 3, 2)
               Dim dif As Decimal = currTotal - Total

               Math.Abs(dif)

               Dim StatisticDiff As Decimal = Math.Round((dif / currTotal) * 100, 2)

               If currTotal > Total Then
                   MsgBox(currTotal & "    : increased by : " & StatisticDiff)
               ElseIf currTotal < Total Then
                   MsgBox(currTotal & "    : decreased by : " & StatisticDiff)
               Else
                   MsgBox(currTotal & "    : stayed the same : " & StatisticDiff)
               End If

Posted 3-Oct-12 23:53pm

Member 9374423

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Kenneth Haugland 4-Oct-12 6:00am

If number < 0 then number*=-1 end if?

3 solutions

Solution 2

Maths.Abs returns a decimal, it doesn't set the value of the supplied parameter.

So, dif = Math.Abs(dif) should fix your problem.

Posted 4-Oct-12 0:02am

DaveAuld

Solution 3

Math.Abs

Method[^] returns an absolute value that contains the result.
You need to capture the return value.

Posted 4-Oct-12 0:02am

Abhinav S

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Joan M · Accepted Answer · 2012-10-04T00:00:00

Solution 1

That is quite normal, you should modify your abs line to:

VB

dif = Math.Abs(dif)

Doing that you'll put the absolute value of dif into dif.

See this link[^].

The Math.Abs function returns the absolute value of a number passed as parameter.
In that way you can do plenty of things:

VB

dif = Math.Abs(dif)
another_var = Math.Abs(dif)
var3 = Math.Abs(function_call())
...

Good luck!

Posted 4-Oct-12 0:00am

Joan M

Updated 4-Oct-12 0:03am

v2

Comments

Member 9374423 4-Oct-12 6:02am

Perfect thank you

Joan M 4-Oct-12 6:03am

you are welcome! glad to help.

Kenneth Haugland 4-Oct-12 6:40am

Thanks, Now I feel stupid... Like a Function instead of a Sub. String.Replace is like that too...

Joan M 4-Oct-12 6:46am

You shouldn't.
I must say that your method:
A) works
B) I've thought of it before posting and searching google about the ABS function.
;)