Very simple.
I want to be able to convert scientific notation to a double in Visual Basic 6.
For example :
5.97427068015303E-10
I have tried :
results.Text = CDbl("5.97427068015303E-10")
results.Text = CDbl(5.97427068015303E-10)
Not working...
The value 5.97427068015303E-10 already is a Double value. I don't think the word convert means what you think it does. Conversion functions certainly won't accomplish what you appear to be after.
I suspect what you are trying to express is that you want to format such tiny values as text in simple decimal notation. If so you might try something like:
Text1.Text = Format$(5.97427068015303E-10, "0.0########################")
Related
i want to display date value like '2020/03/30' as '20200330'
i tried this code
Date.Now.Year & Date.Now.Month & Date.Now.day
but this returns 2020330
how can this return 20200330 ??
You are concatenating numbers, which get implicitly converted to strings.
You could check out Custom date and time format strings.
In your case, this might do just fine: Date.Now.ToString("yyyyMMdd")
By the way, as #jmcilhinney already commented to your question, you could easily find such basic information by using your favorite search engine on the internet...
I have a string with commas in between. How should I convert this string into an integer. I tried using
x?number
but that gives me the following error
Exceptionfreemarker.core.NonNumericalException
e.g. The string is "453,000". I need to convert this to 453000.
Is there any other way of doing this?
There's no function built in for parsing numbers with national formats. ?number only deals with computer format, because when numbers are transferred as strings (which should be already rare), that's what used to be used. So in principle x should be already a number when it gets to FreeMarker, or at least it should use computer format. If that's not possible, you will need a custom function (or method) for that.
I have a rectangular Coordinate in Matlab that looks like the following:
0.0240 - 0.1680i
I'd like to split the double into it real and imaginary parts, those parts being 0.0240 and -0.1680 (Don't need the i here)
I've converted the double into a string using the following:
I=0.0240 - 0.1680*i
I_1=num2str(I)
Im not sure how to proceed here to get what i want. strsplit() just gives back the string in the form it already is. Id like to somehow split it to give me the two numbers separately. I'm not too experienced with data manipulation in Matlab so any help is appreciated.
num2str converts number to string. It is not for separating real and imaginary parts.
You can use:
I=0.0240 - 0.1680*i;
real_part=real (I)
imaginary_part=imag(I)
I want to remove all the characters from a string expect whatever character is between a certain set of characters. So for example I have the input of Grade:2/2014-2015 and I want the output of just the grade, 2.
I'm thinking that I need to use the FIND function to grab whatever is between the : and the / , this also needs to work with double characters such 10 however I believe that it would work so long as the defining values with the FIND function are correct.
Unfortunately I am totally lost on this when using the FIND function however if there is another function that would work better I could probably figure it out myself if I knew what function.
It's not particularly elegant but =MID(A1,FIND(":",A1)+1,FIND("/",A1) - FIND(":",A1) - 1) would work.
MID takes start and length,FIND returns the index of a given character.
Edit:
As pointed out, "Grade:" is fixed length so the following would work just as well:
=MID(A1,7,FIND("/",A1) - 7)
You could use LEFT() to remove "Grade:"
And then use and then use LEFTB() to remove the year.
Look at this link here. This is the way I would go about it.
=SUBSTITUTE(SUBSTITUTE(C4, "Grade:", ""), "/2014-2015", "")
where C4 is the name of your cell.
I have a little Problem in Blitzmax.
I try to read an INI-file and if I read floats they are converted in a very strange way.
The line in the file which is concerned looks like that for example:
_fStrength=40.6
The Output of this looks like that:
DebugLog:_fStrength: 40.5999985
The code I use to read that works with reflection and looks like that:
For Local fld:TField = EachIn id.EnumFields()
fld.Set(obj, SearchInFile("TempWeapon" + index, fld.Name(), "Weapons.ini"))
DebugLog(fld.Name() + ": " + String(fld.Get(obj)))
Next
I found out, that this only happens if the number after the "." does not equal's 5 or 0.
I can't explain this behaviour, because if I do not use reflections, it works fine.
Could anyone help me please?
As you probably know, your computer stores numbers in binary code, using a limited size. 40.6 expanded in binary is a periodic sequence (101000.1001100110011001100..., infinitely), similarly to what happens when you try to write down the digits of 1/3) repeating and thus can not be represented exactly, so you get rounding errors.
The number of correct digits you get here looks like you are using single-precision floating point numbers, you can push the error further back by going to double, but it won't disappear.
As a reference, you might find Wikipedia on floating point helpful.