Kotlin how to make 123.456 format - kotlin

I have scales application. I want to display i.e. 123.456g
I have uint8 that represents 123 (scales range is 0 - 150g)
I also have uint16 that represents 456 (scales range .0 - .999g)
My code:
val strValue = characteristic.getIntValue(FORMAT_UINT16,0).toFloat() + ((characteristic.getIntValue(FORMAT_UINT16,1) or characteristic.getIntValue(FORMAT_UINT16,2).shl(8)) / 1000f)
Produces 51323.457 instead of 123.456
What is wrong?

You're adding floats, not generating a string.
Convert your digits to String, then concatenate them.

Related

Formatting a double variable using String Format to add up to 3 zero from the right

I got a price decimal which sometimes can be either 0.00002001 or 0.00002.
I want to display always 3 zeros from the right if the number is like 0.00002 so I'm looking it to be 0.00002000. If the number is 0.00002001 do not add anything.
I came accross some examples and other examplesin msdn and tried with
price.ToString.Format("{0:F4}", price)
but It doesn't actually change anything in the number.
And in the case number is like 123456789 I want it to display 123.456.789 which I've half solved using ToString("N2") but it's displaying also a .00 decimals which I don't want.
Some special cases here between the fractional and whole numbers, so they need to be handled differently.
Private Function formatWithTrailingZeros(number As Double) As String
If number Mod 1 > 0 Then ' has a fractional component
Return $"{number:0.00000000}"
Else
Dim formattedString = $"{number:N2}"
Return formattedString.Substring(0, formattedString.Length - 3)
End If
End Function
Dim price = 0.00002001
Console.WriteLine(formatWithTrailingZeros(price))
price = 0.00002
Console.WriteLine(formatWithTrailingZeros(price))
price = 123456789
Console.WriteLine(formatWithTrailingZeros(price))
price = 123456789.012345
Console.WriteLine(formatWithTrailingZeros(price))
0.00002001
0.00002000
123,456,789
123456789.01234500
If your second case with 123.456.789 is not based on your current culture, then you may need to replace , with . such as
Return formattedString.Substring(0, formattedString.Length - 3).Replace(",", ".")
Since you are using . both as a decimal separator and a thousands separator, I'm not sure how my example of 123456789.012345000 should look, but since you didn't ask, I'm not going to guess.

FormatNumber replacing number with 0

Not understanding this:
Number returned from DataReader: 185549633.66000035
We have a requirement to maintain the number of decimal places per a User Choice.
For example: maintain 7 places.
We are using:
FormatNumber(dr.Item("Field"), 7, TriState.false, , TriState.True)
The result is: 185,549,633.6600000.
We would like to maintain the 3 (or 35) at the end.
When subtracting two numbers from the resulting query we are getting a delta but trying to show these two numbers out to 6,7,8 digits is not working thus indicating a false delta to the user.
Any advice would be appreciated.
Based on my testing, you must be working with Double values rather than Decimal. Not surprisingly, the solution to your problem can be found in the documentation.
For a start, you should not be using FormatNumber. We're not in VB6 anymore ToTo. To format a number in VB.NET, call ToString on that number. I tested this:
Dim dbl = 185549633.66000035R
Dim dec = 185549633.66000035D
Dim dblString = dbl.ToString("n7")
Dim decString = dec.ToString("n7")
Console.WriteLine(dblString)
Console.WriteLine(decString)
and I saw the behaviour you describe, i.e. the output was:
185,549,633.6600000
185,549,633.6600004
I read the documentation for the Double.ToString method (note that FormatNumber would be calling ToString internally) and this is what it says:
By default, the return value only contains 15 digits of precision although a maximum of 17 digits is maintained internally. If the value of this instance has greater than 15 digits, ToString returns PositiveInfinitySymbol or NegativeInfinitySymbol instead of the expected number. If you require more precision, specify format with the "G17" format specification, which always returns 17 digits of precision, or "R", which returns 15 digits if the number can be represented with that precision or 17 digits if the number can only be represented with maximum precision.
I then tested this:
Dim dbl = 185549633.66000035R
Dim dblString16 = dbl.ToString("G16")
Dim dblString17 = dbl.ToString("G17")
Console.WriteLine(dblString16)
Console.WriteLine(dblString17)
and the result was:
185549633.6600004
185549633.66000035

How to format integer as string with 2 digits?

I would like to format an integer 9 to "09" and 25 to "25".
How can this be done?
You can use either of these options:
The "0" Custom Specifier
value.ToString("00")
String.Format("{0:00}", value)
The Decimal ("D") Standard Format Specifier
value.ToString("D2")
String.Format("{0:D2}", value)
For more information:
Custom Numeric Format Strings
Standard Numeric Format Strings
If its just leading zero's that you want, you can use this:
value.tostring.padleft("0",2)
value.ToString().PadLeft(2, '0'); // C#
If you have 2 digits, say 25 for example, you will get "25" back....if you have just one digit, say 9 for example, you will get "09"....It is worth noting that this gives you a string back, and not an integer, so you may need to cast this later on in your code.
String formate is the best way to do that. It's will only add leading zero for a single length. 9 to "09" and 25 to "25".
String.format("%02d", value)
Bonus:
If you want to add multiple leading zero 9 to "0009" and 1000 to "1000". That's means you want a string for 4 indexes so the condition will be %04d.
String.format("%04d", value)
I don't know the exact syntax. But in any language, it would look like this.
a = 9
aString =""
if a < 10 then
aString="0" + a
else
aString = "" + a
end if

Converting binary to base 4

What I hope to achieve:
I want to convert text to DNA (which is a base 4 system, "a,G,T,c")
How I plan to do it:
Convert text string to binary,
Dim BinaryConvert As String = ""
For Each C As Char In Textbox1.Text
Dim s As String = System.Convert.ToString(AscW(C), 2).PadLeft(8, "0")
BinaryConvert &= s
Next
Textbox1.Text = BinaryConvert '//Changes the textbox1.Text into binary form
Then convert binary to base 4 via Pseudocode solution:
if (length of binary String is an odd number) add a zero to the front (leftmost position) of the String.
Create an empty String to add translated digits to.
While the original String of binary is not empty {
Translate the first two digits only of the binary String into a base-4 digit, and add this digit to the end (rightmost) index of the new String.
After this, remove the same two digits from the binary string and repeat if it is not empty.
}
The idea behind converting binary to DNA is simply setting G and T equal to one, with c and a equal to zero (G=T=1, a=c=0).
So all I have to do is convert the string to binary first, and then into base 4, in order to convert text to genetic code. Could you please help me write the code to convert binary to base 4.
Thank you for the help!
Converting to base 4 from base 2 is pretty simple. Since 4 itself is the 2nd power of 2, this means you can simply combine two bits to create one base 4 place (2 bits can represent 4 possible values, while 1 base 4 place can also represent 4 possible values). For example:
11100100 (base 2) = 3210 (base 4)

Format a number to display a comma when larger than a thousand

I am writing some code in Visual Basic.net and have a question.
If I have a long number, that is larger than 1000, how can I format this value to be 1,000 (with a comma) and for this to be stored in a string?
For e.g.
1234 will be stored as 1,234
12345 will be stored as 12,345
123456 will be stored as 123,456
Is this done with a TryParse statement?
May I have some help to so this?
Take a look at The Numeric ("N") Format Specifier
General use:
Dim dblValue As Double = -12445.6789
Console.WriteLine(dblValue.ToString("N", CultureInfo.InvariantCulture))
' Displays -12,445.68
If you are only using integers then the following:
Dim numberString As String = 1234.ToString("N0")
Will show numberString = "1,234" as the "N0" format will not add any figures after a decimal point.
For those wanting to do a currency with commas and decimals use the following: .ToString("$0,00.00")
Using $ notation:
int myvar = 12345;
Console.WriteLine($"Here is my number: {myvar:N0}");