I'm new to VB.net coding i would like to know how to round of a decimal number to the nearest integer
Eg. X= (5-2/2) = 1.5
but I need only as 1.
Thank you.
You can use the integer division operator if you just want to discard any remainder:
Dim resultValue As Integer = (5-2) \ 2
Note that this is one of the differences between VB.NET and C#, in C# the normal division operator will always apply integer division, so discard the remainder.
You have other options:
resultValue = CInt(Math.Floor((5-2) / 2))
resultValue = CInt(Math.Round((5-2) / 2, MidpointRounding.ToZero))
Try this:
Math.Round( (5-2)/2, 0)
..and look into the options for the overload with the MidpointRounding param, which you can set to influence rounding when the result ends with .5. Here are some of the available options:
Math.Round((5 - 2) / 2, MidpointRounding.AwayFromZero)
Math.Round((5 - 2) / 2, MidpointRounding.ToEven)
Check the documentation (or trial & error) to see which best suits your needs.
Or if you always want either the integer below or the integer above, then also check out Math.Floor and Math.Ceiling functions.
I think you can use the floor function. For example, double floor(double x);, the floor function returns the largest integer that is smaller than or equal to x.
Related
Vb.net has a decimal data type.
Unlike normal double or floating points, decimal data type can store values like 0.1
Now say I have a variable like precision.
Say precision is 8
So basically I want to do
Protected Overridable Sub setPairsPricesStep2(decimalPrecission As Long, Optional base As String = "", Optional quote As String = "")
If decimalPrecission = 8 Then
Return
End If
Dim price = 10D ^ (-decimalPrecission)
setPairsPriceStep1(price, base, quote)
End Sub
There is a problem there
the result of Dim price = 10D ^ (-decimalPrecission) is double, not decimal. I can convert it to decimal but then I will lost the precission.
So what is the right way to do it? Should I just use for next but that's hardly elegant.
It's simple
I want a function that given precisions give decimal value.
For example, if precision is 1 I got 0.1. If precision is 5, I got 0.00001
I ended up doing this
For i = 1 To decimalPrecission
price *= 0.1D
Next
But surely there is a better way
Update:
Per comment, I tried
Dim e = 10D ^ -5
Dim e1 = 10D ^ -5L
The type of e and e1 are both double.
I suppose I can do Cdec(e). But then it means I have lost accuracy because normal double cannot store .1 correctly.
I want a function that given precisions give decimal value.
For example, if precision is 1 I got 0.1. If precision is 5, I got 0.00001
Since you are working with the Decimal type, the simplest way to get this result is to use the Decimal constructor that allows you to specify the scale factor.
Public Sub New (lo As Integer, mid As Integer, hi As Integer, isNegative As Boolean, scale As Byte)
From the Remarks section of the above referenced documentation,
The binary representation of a Decimal number consists of a 1-bit sign, a 96-bit integer number, and a scaling factor used to divide the integer number and specify what portion of it is a decimal fraction. The scaling factor is implicitly the number 10 raised to an exponent ranging from 0 to 28.
So you can see that if take the value of one divided by 10 to the first power, the result is 0.1. Likewise, one divided by 10 to the fifth power, the result is 0.00001.
The lo, mid, and hi arguments in the constructor could be obtained by uisng the [Decimal.GetBits Method](Decimal.GetBits Method), but for this simple case, I chose to hard code the values for the value of one stored as a decimal.
To obtain a value of 0.1D:
New Decimal(1, 0, 0, False, 1)
To obtain a value of 0.00001D:
New Decimal(1, 0, 0, False, 5)
Dim stringrepresentation = "1E-" + decimalPrecission.ToString
Dim price = Decimal.Parse(stringrepresentation, System.Globalization.NumberStyles.AllowExponent)
This is what I basically did. Basically I created a string 1E-5, for example, and use decimal.parse to get the decimal
I wonder if there is a better way but I have no idea.
Actually Jimy ways may work too but rounded to a number
Problems like this drive me crazy. Here's the relevant piece of code:
Dim RES As New Size(Math.Floor(Math.Round(mPageSize.Width - mMargins.Left - mMargins.Right - mLabelSize.Width, 4) / (mLabelSize.Width + mSpacing.Width) + 1),
Math.Floor((mPageSize.Height - mMargins.Top - mMargins.Bottom - mLabelSize.Height) / (mLabelSize.Height + mSpacing.Height)) + 1)
Values of the variables (all are of Single type):
mPageSize.Width = 8.5
mMargins.Left = 0.18
mMargins.Right = 0.18
mLabelSize.Width = 4.0
mSpacing.Width = 0.14
For God-knows-what reason, RES evaluates to {Width=1,Height=5} instead of {Width=2,Height=5}. I have evaluated the expressions on the right-side individually and as a whole and they correctly evaluate to {2,5}, but RES would never get correct value. Wonder what am I missing here.
EDIT
I have simplified the problem further. The following code will produce 2.0 if you QuickWatch the RHS, but the variable on the LHS will get 1.0 after you execute this line:
Dim X = Math.Floor(Math.Round(mPageSize.Width - mMargins.Left - mMargins.Right - mLabelSize.Width, 4) / (mLabelSize.Width + mSpacing.Width) + 1)
Time for MS to check it out?
EDIT 2
More info. The following gives correct results:
Dim Temp = mPageSize.Width - mMargins.Left - mMargins.Right - mLabelSize.Width
Dim X = Math.Floor(Temp / CDec(mLabelSize.Width + mSpacing.Width)) + 1
The problem is that the following expression evaluates to a value just below 1:
Math.Round(mPageSize.Width - mMargins.Left - mMargins.Right - mLabelSize.Width, 4) / (mLabelSize.Width + mSpacing.Width)
= 0.99999999985602739 (Double)
But what's the reason for that? The truth is that I don't know exactly. The MSDN does not offer enough information about the implementation of / but here's my guess:
Math.Round returns a Double with value 4.14. The right-hand side of the division is a Single. So you're dividing a Double by a Single. This results in a Double (see MSDN). So far, so good. The MSDN states that all integral data types are widened to Double before the division. Although Single is not an integral data type, this is probably what happens. And here is the problem. The widening does not seem to be performed on the result of the addition, but on its operands.
If you write
Dim sum = (mLabelSize.Width + mSpacing.Width) 'will be 4.14 Single
Math.Round(mPageSize.Width - mMargins.Left - mMargins.Right - mLabelSize.Width, 4) / sum
= 1 (Double)
Here sum is converted to double (resulting in 4.14) and everything is fine. But, if we convert both operands to double, then the conversion of 0.14 introduces some floating point error:
Dim dblLabelSizeWidth As Double = mLabelSize.Width ' will be 4.0
Dim dblSpacing As Double = mSpacing.Width ' will be 0.14000000059604645
The sum is slightly bigger than 4.14, resulting in a quotient slightly smaller than 1.
So the reason is that the conversion to double is not performed on the division's operand, but on the operand's operands, which introduces floating point errors.
You could overcome this problem by adding a small epsilon to the quotient before rounding off. Alternatively you might consider using a more precise data type such as Decimal. But at some point, there will also be floating-point errors with Decimal.
This is due to rounding error: you're taking the floor of a value that is very close to 2, but is less than 2 (while the mathematical value is 2). You should do all your computations with integers, or take rounding errors into account before using operations like floor (not always possible if you want the true value).
EDIT: Since vb.net has a Decimal datatype, you can also use it instead of integers. It may help in some cases like here: the base conversions for 0.18 and 0.14 (not representable exactly in binary) are avoided and the additions and subtractions will be performed exactly here, so that the operands of the division will be computed exactly. Thus, if the result of the division is an integer, you'll get it exactly (instead of possibly a value just below, like what you got with binary). But make sure that your inputs are already in decimal.
I've read much about rounding in Excel. I found out that VBA's Round() function uses "Bankers rounding" while Application.WorksheetFunction.Round() uses more or less "normal" rounding. But it didn't help me to understand this:
? Round(6.03499,2)
6.03
Why? I want to see 6.04, not 6.03! The trick is that
? Round(Round(6.03499,3),2)
6.04
I thought a bit and developed a subroutine like this:
Option Explicit
Function DoRound(ByVal value As Double, Optional ByVal numdigits As Integer = 0) As Double
Dim i As Integer
Dim res As Double
res = value
For i = 10 To numdigits Step -1
res = Application.Round(res, i)
Next i
DoRound = res
End Function
It works fine.
? DoRound(6.03499,2)
6.04
But it is not cool. Is there any built-in normal rounding in Excel?
If you round 6.03499 to 3 digits it will be 6.035 - which is correct.
If you round 6.03499 to 2 digits it will be 6.03 - which is correct
However - the example where you first round to 3 digits, then to 2 is also correct, by the following statement:
Round(6.03499, 3) gives 6.035
Round(6.035, 2) gives 6.04
If you want Round(6.03499, 2) to give 6.04 you have to use Application.WorksheetFunction.RoundUp
Rounding 6.0349 to two decimals is just not 6.04 hence, no, there is no such function.
Round up will round anything up. Hence, 6.0000000001 will also become 7 if you round to 0 decimals.
I have code like this:
Dim minutes As Integer = (55 / 60)
I want this to return 0 to me, integer 0 (no decimals), but VB.NET rounds this to 1.
How do I accomplish this?
OK for those with the same problem, try dividing with \ (yeah, it's not escape character in VB.NET).
For full decimal division you have to use another operator.
Try:
\
Dim minutes As Integer = Math.Round(55 / 60)
There are overloads that allow rounding to fractional values and specifying which way to round if the value is exactly between two integers.
So round down? Use:
?Math.Truncate(55/60)
0.0
What is the syntax to round up a decimal leaving two digits after the decimal point?
Example: 2.566666 -> 2.57
If you want regular rounding, you can just use the Math.Round method. If you specifially want to round upwards, you use the Math.Ceiling method:
Dim d As Decimal = 2.566666
Dim r As Decimal = Math.Ceiling(d * 100D) / 100D
Here is how I do it:
Private Function RoundUp(value As Double, decimals As Integer) As Double
Return Math.Ceiling(value * (10 ^ decimals)) / (10 ^ decimals)
End Function
Math.Round is what you're looking for. If you're new to rounding in .NET - you should also look up the difference between AwayFromZero and ToEven rounding. The default of ToEven can sometime take people by surprise.
dim result = Math.Round(2.56666666, 2)
You can use System.Math, specifically Math.Round(), like this:
Math.Round(2.566666, 2)
Math.Round(), as suggested by others, is probably what you want. But the text of your question specifically asked how to "roundup"[sic]. If you always need to round up, regarless of actual value (ie: 2.561111 would still go to 2.57), you can do this:
Math.Ceiling(d * 100)/100D
The basic function for rounding up is Math.Ceiling(d), but the asker specifically wanted to round up after the second decimal place. This would be Math.Ceiling(d * 100) / 100. For example, it may multiply 46.5671 by 100 to get 4656.71, then rounds up to get 4657, then divides by 100 to shift the decimal back 2 places to get 46.57.
I used this way:
Math.Round(d + 0.49D, 2)
Math.Ceiling((14.512555) * 100) / 100
Dot net will give you 14.52. So, you can use above syntax to round the number up for 2 decimal numbers.
I do not understand why people are recommending the incorrect code below:
Dim r As Decimal = Math.Ceiling(d * 100D) / 100D
The correct code to round up should look like this:
Dim r As Double = Math.Ceiling(d)
Math.Ceiling works with data type Double (not Decimal).
The * 100D / 100D is incorrect will break your results for larger numbers.
Math.Ceiling documentation is found here: http://msdn.microsoft.com/en-us/library/zx4t0t48.aspx