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I tried....
Dim DisTest As Double = Me.TermDiscPct / 100
Round(DisTest, 2, MidpointRounding.AwayFromZero)
and it didn't make any difference
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The Round method doesn't change the value that you pass to it, you have to take care of the return value:
DisTest = Round(DisTest, 2, MidpointRounding.AwayFromZero)
Despite everything, the person most likely to be fooling you next is yourself.
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Thanks Guffa, I caught that earlier (stated it further down in the posts).
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Function RoundMyDouble(ByVal r As Double, ByVal places As UInt16)
Dim off As Double = Math.Pow(10, places)
Return CInt(r * off) / off
End Function
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I tried....
Dim DisTest As Double = Me.TermDiscPct / 100
Round(DisTest, 2, MidpointRounding.AwayFromZero)
and it didn't make any difference
I don't want to create a function just for this. It seems like there should be a way to just round this with that much overhead.
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Are you using Math.Round function? 
Just to check because it works fine with me, it returns 0.03
if I use 0.025...
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Yes I was doing that(Math.Rounding) but I wasn't resigning the variable the new value. That works just fine. Thanks!!
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You might want to double check the results and do some testing to see if it's going to work for you. Also, doing math on the number you want to round while using a Double type can affect how rounding works.
For example, Doing something like: Math.Round(1.255, 2) results in 1.25, not 1.26. What happens is that 1.255 is multiplied by 100 (the 2 parameter), then the number is rounded to the nearest integer. But, with the Double type, this results in 1.255 * 100 = 125.4999999999, which will round DOWN to the nearest integer of 125. This is then divided by 100, which results in 1.25. If accuracy is mandatory, consider using the Decimal type. It's slower since it's not a native CPU data type, but it is more accurate.
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Hey Dave, thanks for your reply. Doesn't MidpointRounding.AwayFromZero fix what you are refering to? I was having a similar problem that you descriped and it took care of it.
I was doing this..
Dim DisTest As Double = 2.5 / 100
Round(DisTest, 2)
And my return was .02 instead of .03 like I would expect.
Now I am doing this...
Dim DisTest As Double = 2.5 / 100
Round(DisTest, 2,MidpointRounding.AwayFromZero)
And now my return is .03 just as I would expect it to be. I am only asking because accuracy is important and it seems to be pretty accurate as of now. But are you saying if I had 1.255 / 100 then it wouldn't be accurate?
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CCG3 wrote: Doesn't MidpointRounding.AwayFromZero fix what you are refering to?
No, it doesn't. It just happened to change the behavior of the Round method so THAT particular case worked out properly. The AwayFromZero option rounds numbers up, never down.
CCG3 wrote: And now my return is .03 just as I would expect it to be. I am only asking because accuracy is important and it seems to be pretty accurate as of now. But are you saying if I had 1.255 / 100 then it wouldn't be accurate?
Yep. The only reason why it worked is because the math using Double numbers worked out to nice even binary representations. For example, 1.255 / 100 = 0.012549999999999, not 0.01255. In your case, 2.5 / 100 worked out evenly to 0.025. It's those binary approximations that screws up the rounding. It just happened to work out properly on the example you tried. It will NOT work in all cases.
You'd probably be better off using the Decimal type instead of Double.
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Thanks Dave, I think I understand what you are saying...
on further testing, I ran these lines just like this...
Dim DisTest As Double = 1.255 / 100
MsgBox(DisTest)
DisTest = Round(DisTest, 2, MidpointRounding.AwayFromZero)
MsgBox(DisTest)
Dim DisTest2 As Decimal = 1.255 / 100
MsgBox(DisTest2)
DisTest2 = Round(DisTest2, 2)
MsgBox(DisTest2)
They both gave the same results. (.01)
And you are saying that I would be better off to just use Decimal. The preceeding example is all I would have to change right? Instead of using Double change it to Decimal and leave out the MidpointRounding.AwayFromZero. right?
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Yep. But, only your testing is going to tell you if it's appropriate or not. Remember, there is a performance hit for using the Deicmal type.
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That's because 1.255 / 100 = .01255
Rounding to 2 places *IS* .01
Any suggestions, ideas, or 'constructive criticism' are always welcome.
"There's no such thing as a stupid question, only stupid people." - Mr. Garrison
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Dim DisTest as Double = 2.5/100
DisTest= (Int((DisTest + 0.005) * 100)) / 100
The secret is in the 0.005.
0.001 + 0.005 = 0.006, * 100 = 0.6, int = 0, / 100 = 0.00
0.002 + 0.005 = 0.007, * 100 = 0.7, int = 0, / 100 = 0.00
0.003 + 0.005 = 0.008, * 100 = 0.8, int = 0, / 100 = 0.00
0.004 + 0.005 = 0.009, * 100 = 0.9, int = 0, / 100 = 0.00
0.005 + 0.005 = 0.010, * 100 = 1.0, int = 1, / 100 = 0.01
0.006 + 0.005 = 0.011, * 100 = 1.1, int = 1, / 100 = 0.01
0.007 + 0.005 = 0.012, * 100 = 1.2, int = 1, / 100 = 0.01
0.008 + 0.005 = 0.013, * 100 = 1.3, int = 1, / 100 = 0.01
0.009 + 0.005 = 0.014, * 100 = 1.4, int = 1, / 100 = 0.01
0.010 + 0.005 = 0.015, * 100 = 1.5, int = 1, / 100 = 0.01
etc
Also, I don't use "Round" because I can never remember from language to language how it works (round up? round down? round on .5?)
If I want to round .005 down, .006 up (standard practise), I use 0.004 instead of 0.005
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Hi all.
I am drawing a form and i have changed its property as
form1.borderstyle=none
I hv created my own border and controls to handle form but boundries of form are not bold as it seems in a bordered form.
I want to make boundries of form bold with having same property of form (means borderless form).
Thanks.
Gagan
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Your question seems to be a bit jumbled. You have created a borderless form that allows interaction (minimize, maximize, close) through custom controls. Now, based on what I understand, can you make the question a bit more clear?
Regards,
Thomas Stockwell
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
Visit my Blog
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As i said, i have created a borderless form and use my custom controls to minimize,maximize and close form.
I just want to make boundries of my form bold. (as in a bordered form)
Thanks.
Gagan
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This can be done in a variety of ways. Image background, panel with 3D styles, custom control form borders, etc.
Regards,
Thomas Stockwell
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
Visit my Blog
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Hi Friends, I am unable to Implement This line of Code of Vb6 in Visual Studio means Vb.Net because .Net does not support Pset Method...
picImage.PSet (r, c), RGB(clrr, clrg, clrb)
where r and c are radom points of picture....
Any body plz Guide me...........
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You should make clear what you want to do. what is this picImage.
Is it a picturebox or an image control.
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PIcimage is a picturebox and clrr = red Color,clrg = green color,clrb=blue......
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I am assuming that the picImage is a Bitmap object (at least when transferred to VB.NET).
You should be able to use the SetPixel method:
picImage.SetPixel(r,c,new Color(clrr,clrg,clrb));
Regards,
Thomas Stockwell
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
Visit my Blog
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Dear this line of code giving me this error...
picImage.SetPixel(r,c,new Color(clrr,clrg,clrb));
Error is:
"Type System.drawing.color has no constructs"
and picimage is a bitmap........
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Oops,
Color.FromARGB(clrr,clrg,clrb)
Regards,
Thomas Stockwell
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
Visit my Blog
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