how can I Show only three significant digits for the data volume, i.e. 0,xxx; x,xx; xx,x; xxx; x’xxx etc.
this is my code
Option Strict On
Imports System.Globalization
Module Module1
Public Function BytesToMegabytes(Bytes As Long) As String
'This function gives an estimate to two decimal
'places. For a more precise answer, format to
'more decimal places or just return dblAns
Dim dblAns As Double = (Bytes / 1024) / 1024
Dim ci = New CultureInfo("en-GB")
ci.NumberFormat.NumberDecimalSeparator = "'"
Return dblAns.ToString("###,###,##0.00", ci)
End Function
Sub Main()
Console.WriteLine(BytesToMegabytes(9225936896))
Console.ReadLine()
End Sub
End Module
Outputs:
currently I score 8,798'54 MB.
to be 8'798 MB, how can I get it?
thank you all for your help
Following on from your previous question...
To use significant figures instead of decimal places:
Public Function BytesToMegabytes(bytes As Long) As String
Dim dblAns As Double = (bytes / 1024) / 1024
If dblAns = 0 Then
Return "0"
End If
Dim significantFigures = 3
Dim magnitude = Math.Floor(Math.Log10(dblAns))
Dim v As Double = 10 ^ (magnitude - significantFigures + 1)
dblAns = Math.Floor(dblAns / v) * v
Dim ci = New CultureInfo("")
ci.NumberFormat.NumberDecimalSeparator = ","
ci.NumberFormat.NumberGroupSeparator = "'"
Return dblAns.ToString("#,##0.###", ci)
End Function
For example, Console.WriteLine(BytesToMegabytes(9225936896)) outputs
8'790
If you change the line Dim significantFigures = 3 to Dim significantFigures = 4, it outputs
8'798
Depending on the rounding you want, you may want to use dblAns = Math.Round(dblAns / v, MidpointRounding.AwayFromZero) * v instead, or perhaps Math.Ceiling instead of Math.Floor.
Related
I'm trying to make a function that takes a three digit number and reverses it (543 into 345)
I can't take that value from a TextBox because I need it to use the three numbers trick to find a value.
RVal = ReverseDigits(Val)
Diff = Val - RVal
RDiff = ReverseDigits(Diff)
OVal = Diff + RDiff
543-345=198
198+891=1089
Then it puts 1089 in a TextBox
Function ReverseDigits(ByVal Value As Integer) As Integer
' Take input as abc
' Output is (c * 100 + b * 10 + a) = cba
Dim ReturnValue As Boolean = True
Dim Val As String = CStr(InputTextBox.Text)
Dim a As Char = Val(0)
Dim b As Char = Val(1)
Dim c As Char = Val(2)
Value = (c * 100) + (b * 10) + (a)
Return ReturnValue
End Function
I've tried this but can't figure out why it won't work.
You can convert the integer to a string, reverse the string, then convert back to an integer. You may want to enforce the three digit requirement. You can validate the argument before attempting conversion
Public Function ReverseDigits(value As Integer) As Integer
If Not (value > 99 AndAlso value < 1000) Then Throw New ArgumentException("value")
Return Integer.Parse(New String(value.ToString().Reverse().ToArray()))
End Function
My code is pretty simple and will also work for numbers that don't have three digits assuming you remove that validation. To see what's wrong with your code, there are a couple of things. See the commented lines which I changed. The main issue is using Val as a variable name, then trying to index the string like Val(0). Val is a built in function to vb.net and the compiler may interpret Val(0) as a function instead of indexing a string.
Function ReverseDigits(ByVal Value As Integer) As Integer
' Dim ReturnValue As Boolean = True
' Dim Val As String = CStr(InputTextBox.Text)
Dim s As String = CStr(Value)
Dim a As Char = s(0)
Dim b As Char = s(1)
Dim c As Char = s(2)
Value = Val(c) * 100 + Val(b) * 10 + Val(a)
'Return ReturnValue
Return Value
End Function
(Or the reduced version of your function, but I would still not hard-code the indices because it's limiting your function from expanding to more or less than 3 digits)
Public Function ReverseDigits(Value As Integer) As Integer
Dim s = CStr(Value)
Return 100 * Val(s(2)) + 10 * Val(s(1)) + Val(s(0))
End Function
And you could call the function like this
Dim inputString = InputTextBox.Text
Dim inputNumber = Integer.Parse(inputString)
Dim reversedNumber = ReverseDigits(inputNumber)
Bonus: If you really want to use use math to find the reversed number, here is a version which works for any number of digits
Public Function ReverseDigits(value As Integer) As Integer
Dim s = CStr(value)
Dim result As Integer
For i = 0 To Len(s) - 1
result += CInt(Val(s(i)) * (10 ^ i))
Next
Return result
End Function
Here's a method I wrote recently when someone else posted basically the same question elsewhere, probably doing the same homework:
Private Function ReverseNumber(input As Integer) As Integer
Dim output = 0
Do Until input = 0
output = output * 10 + input Mod 10
input = input \ 10
Loop
Return output
End Function
That will work on a number of any length.
I am writing a vb.net program to generate a three digit serial number I will use in printing a barcode.
The requirements are the counter must count:
001 - 999, A00 - A99, B00 - B99, ..., Z00 - Z99
I cannot use the letters O and I
This code simply increments the value I pass to it by 1. I first check if the value is <=998 and if so return the value in 3 digits. I had to put this in a Try statement because passing the value 'A00' caused an error.
The code is still breaking once I hit Z99.
Problem: If the next serial number = Z90 and the user wants to print 35 barcodes I need to stop the operation before it begins and warn the user there are only 10 avail serial numbers remaining
Also, I am also hoping for advice on how I could have accomplished this in a better manner, any advice would be greatly appreciated
Public Shared Function NextSerial(ByVal value As String) As String
Try
If value <= 998 Then
value += 1
Return ZeroPad(value, 3)
End If
Catch ex As Exception
End Try
Const chars As String = "ABCDEFGHJKLMNPQRSTUVWXYZ"
Dim threenumber As String = ZeroPad(value, 3) 'ensure value is 3 digits.
Dim alpha As String = threenumber.Substring(0, 1).ToUpper() ' 1st digit
Dim beta As String = threenumber.Substring(1, 2) 'remaining two digits
Dim newNumber As String
Dim nextletter As String
If beta = "99" Then
beta = "00"
nextletter = chars.Substring((chars.IndexOf(alpha, System.StringComparison.Ordinal) + 1), 1)
newNumber = nextletter + beta
Return newNumber
Else
beta += 1
newNumber = alpha + ZeroPad(beta, 2)
Return newNumber
End If
End Function
Private Shared Function ZeroPad(ByVal number As String, ByVal toLength As Integer) As String
ZeroPad = number
'add the necessary leading zeroes to build it up to the desired length.
Do Until Len(ZeroPad) >= toLength
ZeroPad = "0" & ZeroPad
Loop
End Function
I think you can do this by assuming your first character is the 'hundreds' and converting to a number and incrementing:
Private Function NextSerial(value As String) As String
Const chars As String = "0123456789ABCDEFGHJKLMNPQRSTUVWXYZ"
Dim numericValue As Integer = 100 * (chars.IndexOf(value.Substring(0, 1))) + Integer.Parse(value.Substring(1, 2))
numericValue += 1
Return chars.Substring(numericValue \ 100, 1) + (numericValue Mod 100).ToString.PadLeft(2, "0")
End Function
You should of course perform some error checking at the start of the function to make sure a valid serial number has been handed into the function. I would also put this function into a class and add functions such as isValid, SerialsRemaining and perhaps a function to retrieve a list of multiple serials.
I created constant strings that represent every available character in each digit position. I then used indexing to lookup the positions of the current serial number & moved one number forward to get the next serial. This will always provide the next serial until you run out of numbers.
Note: this code can easily be made more compact, but I left it as-is thinking it might be clearer.
Const charString1 As String = "0123456789ABCDEFGHJKLMNPQRSTUVWXYZ"
Const charString2 As String = "0123456789"
Const charString3 As String = "0123456789"
Public Function NextSerial(ByVal value As String) As String
' ensures the input is three chars long
Dim threenumber As String = Right("000" & value, 3)
Dim char1 As String = threenumber.Substring(0, 1)
Dim char2 As String = threenumber.Substring(1, 1)
Dim char3 As String = threenumber.Substring(2, 1)
Dim char1Pos As Integer = charString1.IndexOf(char1)
Dim char2Pos As Integer = charString2.IndexOf(char2)
Dim char3Pos As Integer = charString3.IndexOf(char3)
If char1Pos = -1 Or char2Pos = -1 Or char3Pos = -1 Then Throw New Exception("Invalid serial number format")
' move to next serial number
char3Pos += 1
If char3Pos > charString3.Length() - 1 Then
char3Pos = 0
char2Pos += 1
End If
If char2Pos > charString2.Length() - 1 Then
char2Pos = 0
char1Pos += 1
End If
If char1Pos > charString1.Length() - 1 Then Throw New Exception("Out of serial numbers!")
Return charString1.Substring(char1Pos, 1) & charString2.Substring(char2Pos, 1) & charString3.Substring(char3Pos, 1)
End Function
I suggest you use integer for all your check and calculation and only convert to serial number for display. It'll be a lot easier to know how many serial number are remaining.
Your serial number is similar to integer except everything over 100 is a letter instead of a number.
Note: It's very important to add error checking, this assumes that all input are valid.
Module Module1
Sub Main()
Console.WriteLine(SerialNumber.ConvertSerialNumberToInteger("D22"))
Console.WriteLine(SerialNumber.ConvertIntegerToSerialNumber(322))
Console.WriteLine(SerialNumber.GetAvailableSerialNumber("Z90"))
For Each sn As String In SerialNumber.GetNextSerialNumber("X97", 5)
Console.WriteLine(sn)
Next
Console.ReadLine()
End Sub
End Module
Class SerialNumber
Private Const _firstPart As String = "ABCDEFGHJKLMNPQRSTUVWXYZ"
Public Shared Function ConvertSerialNumberToInteger(ByVal serialNumber As String) As Integer
Return (_firstPart.IndexOf(serialNumber(0)) * 100) + Integer.Parse(serialNumber.Substring(1, 2))
End Function
Public Shared Function ConvertIntegerToSerialNumber(ByVal value As Integer) As String
Return _firstPart(value \ 100) & (value Mod 100).ToString("00")
End Function
Public Shared Function GetAvailableSerialNumber(ByVal serialNumber As String)
Dim currentPosition As Integer
Dim lastPosition As Integer
currentPosition = ConvertSerialNumberToInteger(serialNumber)
lastPosition = ConvertSerialNumberToInteger("Z99")
Return lastPosition - currentPosition
End Function
Public Shared Function GetNextSerialNumber(ByVal serialNumber As String, ByVal amount As Integer) As List(Of String)
Dim newSerialNumbers As New List(Of String)
Dim currentPosition As Integer
currentPosition = ConvertSerialNumberToInteger(serialNumber)
For i As Integer = 1 To amount
newSerialNumbers.Add(ConvertIntegerToSerialNumber(currentPosition + i))
Next
Return newSerialNumbers
End Function
End Class
I recently wrote this program in Visul Basic 13.
it searchs for the nth catalan number but after 48 even Decimal type is too short.
Is there any other way to represent them? like in the form of A*10^n?
Public Class Try_Catalan_Number
'Catalan numbers form a sequence of natural numbers that occur in various counting problems,
'often involving recursively defined objects.
Inherits Base_Number
Public Overrides Sub Test()
Dim Return_Catalan_Value As Decimal
If Function_Catalan(Return_Catalan_Value) = False Then
Return_To_Form_Boolean = False
Else
Return_To_Form_Boolean = True
End If
Return_To_Form_Value = Function_Catalan(Return_Catalan_Value)
End Sub
Private Function Function_Catalan(Return_Catalan_Value As Decimal) As Decimal
'We return a Decimal function because catalan numbers can be very big and decimal is the biggest type.
Dim Binomial_Cofficients As Decimal
Dim Result As Decimal
Dim Number_Of_Loops As Integer
Dim tmpNumber As Object
Dim K As Decimal
Dim N As Decimal
If (Number > 48) Then
Return False
Exit Function
End If
'48 is the largest catalan number position which can be displayed...any position above 48 is too big.
tmpNumber = Number - 1
N = 2 * tmpNumber
K = tmpNumber
Result = 1
For Number_Of_Loops = 1 To K
Result = Result * (N - (K - Number_Of_Loops))
Result = Result / Number_Of_Loops
Next Number_Of_Loops
Binomial_Cofficients = Result
tmpNumber = Number - 1
tmpNumber = ((1 / (1 + tmpNumber)) * Binomial_Cofficients)
Return_Catalan_Value = tmpNumber
Return Return_Catalan_Value
End Function
End Class
[I assume by "Visul Basic 13" you mean the VB which is associated with Visual Studio 2013, i.e. VB version 12.0.]
You can use System.Numerics.BigInteger (you'll have to add a reference to System.Numerics):
Imports System.Numerics
Module Module1
Friend Function Factorial(n As Integer) As BigInteger
If n < 2 Then Return 1
If n = 2 Then Return 2
Dim f As BigInteger = BigInteger.Parse("2")
For i = 3 To n
f *= i
Next
Return f
End Function
Friend Function CatalanNumber(n As Integer) As BigInteger
Return Factorial(2 * n) / (Factorial(n + 1) * Factorial(n))
End Function
Sub Main()
For i = 0 To 550
Console.WriteLine(CatalanNumber(i).ToString())
Next
Console.ReadLine()
End Sub
End Module
I did not test to see the maximum Catalan number it can calculate, and I have no inclination to verify the results beyond those shown on the Wikipedia page.
Optimisations are left as an exercise for the reader ;)
Edit: FWIW, I can get it to run a bit faster by using
Function CatalanNumber(n As Integer) As BigInteger
Dim nFactorial = Factorial(n)
Dim twonFactorial = nFactorial
For i = (n + 1) To 2 * n
twonFactorial = BigInteger.Multiply(twonFactorial, i)
Next
Return twonFactorial / (BigInteger.Pow(nFactorial, 2) * (n + 1))
End Function
The speed increase varies from roughly 50% (n=50) to 20% (n=5000). If you're only using the function a few times for fairly small n, there may be little point worrying about it.
Edit2 Re-writing your function a bit to make it easier to read and removing the off-by-one error, we get:
Private Function Function_Catalan(a As Integer) As BigInteger
If a = 0 Then Return 1
Dim binomialCofficient As BigInteger = BigInteger.One
Dim n As Integer = 2 * a
Dim k As Integer = a - 1
For i As Integer = 1 To k
binomialCofficient = binomialCofficient * (n - (k - i)) / i
Next i
Return binomialCofficient / a
End Function
to get this format you could use:
String.Format("{0:E4}", InputNumber)
I have a VB.Net application that goes through a series of processes to decode a string, one problem with this is that I have found a function that converts from binary to decimal for any number, but I cannot find a function that will convert a supplied number (in string format) into a binary string. For reference, the binary to decimal conversion function is below:
Public Function baseconv(d As String)
Dim N As Long
Dim Res As Long
For N = Len(d) To 1 Step -1
Res = Res + ((2 ^ (Len(d) - N)) * CLng(Mid(d, N, 1)))
Next N
Return Str(Res)
End Function
What i would do if the number is less than 16*1e18 <-> can hold in a Uint64 :
1) store the number inside an unsigned 64 bits integer : num.
2) then just loop on each bit to test it :
mask = 1<<63
do
if ( num AND mask ) then ->> push("1")
else ->> push("0")
mask = mask >> 1
until mask = 0
(where push builds the output with a string concatenation, or, if performance matters, a StringBuilder, or it can be a stream,... )
what about this?
Module Module1
Sub Main()
Console.WriteLine(Convert.ToString(2253483438943167 * 5, 2))
Console.ReadKey()
End Sub
End Module
Try using the System.Numerics.BigInteger class like this:
Dim d As String
d = "2253483438943167"
Dim bi As BigInteger
bi = BigInteger.Parse(d)
Dim ba() As Byte
ba = bi.ToByteArray()
Dim s As String
Dim N As Long
Dim pad as Char
pad = "0"c
For N = Len(ba) To 1 Step -1
s = s + Convert.ToString(ba(N)).PadLeft(8, pad)
Next N
How about
Dim foo As BigInteger = Long.MaxValue
foo += Long.MaxValue
foo += 2
Dim s As New System.Text.StringBuilder
For Each b As Byte In foo.ToByteArray.Reverse
s.Append(Convert.ToString(b, 2).PadLeft(8, "0"c))
Next
Debug.WriteLine(s.ToString.TrimStart("0"c))
'10000000000000000000000000000000000000000000000000000000000000000
i have a sub that will try to get random foods in the database, get the sum of these food's calories and check if they wont exceed the required calories.
most of the time the sub worked, but sometimes this error appears.
this is my code. it is kinda long.
Private Sub lunchgenerate()
Dim grams As Integer = DSgrams.Tables("grams").Rows(0).Item("grams")
Dim grams1 As Integer = DSricegrams.Tables("ricegrams").Rows(0).Item("grams")
Dim grams2 As Integer = DSgrams2.Tables("grams").Rows(0).Item("grams")
Dim calorieval As Decimal = foodcalories * grams
Dim calorieval1 As Decimal = ricecalories * grams1
Dim calorieval2 As Decimal = foodcalories2 * grams2
Dim carbval As Decimal = foodcarb * grams
Dim carbval1 As Decimal = ricecarb * grams1
Dim carbval2 As Decimal = foodcarb2 * grams2
Dim proteinval As Decimal = foodprotein * grams
Dim proteinval1 As Decimal = riceprotein * grams1
Dim proteinval2 As Decimal = foodprotein2 * grams
Dim fatval As Decimal = foodfat * grams
Dim fatval1 As Decimal = ricefat * grams1
Dim fatval2 As Decimal = foodfat2 * grams
Dim caloriepercent As Decimal = usercalories * 0.5
Dim mincalories As Decimal = caloriepercent - 300
Dim proteinpercernt As Decimal = userprotein * 0.5
Dim minprotein As Decimal = proteinpercernt - 20
Dim counter As Integer = 0
Dim counter1 As Integer = 0
Dim foodcalorietotal As Decimal = calorieval + calorieval1 + calorieval2
Dim foodproteintotal As Decimal = proteinval + proteinval1 + proteinval2
Dim carbtotal As Decimal = carbval + carbval1 + carbval2
Dim foodfattotal As Decimal = fatval + fatval1 + fatval2
If foodcalorietotal < mincalories Or foodcalorietotal > caloriepercent Then
counter = 0
Else
counter = 1
End If
If foodproteintotal < minprotein Or foodproteintotal > proteinpercernt Then
counter1 = 0
Else
counter1 = 1
End If
If counter = 1 And counter1 = 1 Then
'output to the form
Else
**lunchgenerate()**
End If
End If
End Sub
i think the error is when the lunchgenerate() sub is called again.
but just like what i said, most of the times it worked, but sometimes it freezes and then this error appears and then highlights the first line of my code which is
Dim DAlunchcategory As New SqlDataAdapter(lunchquery, CN)
Dim DSlunchcategory As New DataSet
DAlunchcategory.Fill(DSlunchcategory, "category_id")
Stack Overflow error happens when there is no space on the stack (part of the process memory used for local variables, parameter values, function results and a few other little things).
Often the code that produces this error does it by executing infinite recursion. If the code that you showed is complete lunchgenerate function, in the case when either counter or counter1 is not 1, lunchgenerate will be called again with exactly the same outcome, and again, and again, until it completely depletes space on the stack and exception is thrown. You need to have some escape logic to avoid it.