I have this integer. A = 100002 of 6 digits and i want to add 2 extra 0 in the middle so it can be a integer of 8 digits.
Result = 10000002
how can i do it?
You can split your number up into two parts, the left and the right side. Then add the zeros to the left side and put back the right side.
100002 -> 100 [left side] 002 [right side]
Dim number As Integer = 100002
Dim rightSide As Integer = number Mod 1000
Dim leftSide As Integer = number - rightSide
leftSide *= 100 ' Add zeros
Dim newNumber As Integer = leftSide + rightSide
With this, 123456 will become 12300456.
There are lots of different ways to answer this, meaning that your question is likely not specific enough to get the answer you want. For example, the_lotus's answer is of course correct (and a more practical solution, too). However, this solution will also yield the result you specify in the simplest possible manner, by subtracting 2, multiplying by 100, and adding 2 again:
Result = (A - 2) * 100 + 2
Since both of these very different methods solve the problem you have posed, it follows that you might want to pose the problem a bit more carefully. For example, if you want to work with numbers other than 100002 (which you haven't said that you do), this solution of course won't allow that. If you want a solution that applies to numbers with other than six digits, the_lotus's solution won't allow that in all cases, either.
Related
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.
I have 16 different options in my program and i have a 16 character variable which is filled with 1's or 0's depending on the options that are selected (0000000000000000 means nothing is selected, 0010101010000101 means options 3,5,7,9,14 and 16 are selected, 1111111111111111 means everything is selected.)
When i run my program, the code looks (using an if statement) for a 1 in the designated character of the 16 digit number and if there is one there then it runs the code for that option, otherwise it skips it..
e.g option 3 looks too see if the 3rd character (0010000000000000) is a 1 and if it is it runs the code.
Now what i am trying to do is generate a list of every different combination that is possible so I can create an option for it to just loop through and run every possible option:
0000000000000001
0000000000000010
0000000000000011
...
1111111111111100
1111111111111110
1111111111111111
I have tried this but i think it may take a couple of years to run jaja:
Dim binString As String
Dim binNUM As Decimal = "0.0000000000000001"
Do Until binNUM = 0.11111111111111111
binString = binNUM.ToString
If binString.Contains(1) Then
If binString.Contains(2) Or binString.Contains(3) Or binString.Contains(4) Or binString.Contains(5) Or binString.Contains(6) Or binString.Contains(7) Or binString.Contains(8) Or binString.Contains(9) Then
Else
Debug.Print(binNUM)
End If
End If
binNUM = binNUM + 0.0000000000000001
After the code above is complete i would then take the output list and remove any instances of "0." and then any lines which had fewer than 16 chararcters (because the final character would be a 0 and not show) I would add a 0 until there was 16 characters. I know this bit might be stupid but its as far a ive got
Is there a faster way I can I generate a list like this in VB.net?
You should be able to get the list by using Convert.ToString as follows:
Dim sb As New System.Text.StringBuilder
For i As Integer = 0 To 65535
sb.AppendLine(Convert.ToString(i, 2).PadLeft(16, "0"c))
Next
Debug.Print(sb.ToString())
BTW: This should finish in under one second, depending on your system ;-)
Create an enum with FlagAttributes, which allows you to do the key functions you list. Here is an example of setting it up in a small project I am working on:
<FlagsAttribute>
Public Enum MyFlags As Integer
None = 0
One = 1
Two = 2
Three = 4
Four = 8
Five = 16
Recon = 32
Saboteur = 64
Mine = 128
Headquarters = 256
End Enum
e.g.
Dim temp as MyFlags
Dim doesIt as Boolean
temp = MyFlags.One
doesIt = temp.HasFlag(MyFlags.Two)
temp = temp OR MyFlags.Three
'etc.
The real advantage is how it prints out, if you want something other than 0, 1 and is much more human friendly.
How do I make a biased random number generator (RNG) in VB.NET?
I know I could make it by fiddling with the output of the Randomize()/Rnd methods, but is there a built-in way of doing this?
I want the biased RNG to give me either a 2 or 4 (though using 1 or 2 as a substitute is also OK by me), with 2 occurring on average 90% of the time and 4 occurring on average 10% of the time.
Create a random number generator to return values from 1-10, if the value from the random number generator is between 1 and 9 send a 2 if the value is 10 send a 4.
You might want to look at this
http://msdn.microsoft.com/en-us/library/vstudio/ctssatww(v=vs.100).aspx?cs-save-lang=1&cs-lang=vb#code-snippet-2
If you want to come out with a mask to generate your values
Here is what I think you can do.
Dim numbers() as integer = {2,2,2,2,4,2,2,2,2,2} ' set 10% for 4, 90% for 2
Dim r as new Random()
Return numbers(r.Next(0, 10))
I tried to search this problem and unlucky to solve.
I have a generated report using the rdlc, now, I want to sum all the positive numbers not including the negative in rdlc coding way.
Credit
------
3
3
3
3
3
3
2
-3
-2
------
Total: 20
So, the main point is ignore the negative and sum the positive numbers. so far this what I have tried and this not the solution of my problem.
=IIF(Fields!creditUnit.Value > 0, Sum(Fields!creditUnit.Value), 0)
Anybody can help me?
UPDATE:
This was my temporary solution, since my generated report is good for only one page. I create a parameter for Total
var total = creditsList.Where(c => c.HasValue && c.credit > 0).Sum(c => c.credit.Value);
var totalParam = new ReportParameter("total", total);
I hope that one of you guys can help me in what would be the solution in rdlc coding way to sum all the positive numbers.
UPDATE:
I included the vb.net because the way the coding of rldc is VB syntax.
try
=Sum(IIF(Fields!creditUnit.Value > 0, Fields!creditUnit.Value, 0))
I know its to late but this will work for me
if column have only integer then following expression can be use.
=Sum(IIf((Fields!creditUnit.Value>0), CInt(Fields!creditUnit.Value), 0))
But for decimal or integer and decimal (both) following expression can be use.
=Sum(IIf((Fields!creditUnit.Value>0),CDbl(Fields!creditUnit.Value),0.0))
I'm writing a task in pascal.
Everything is ok, just my result is not right.
I'm summing some numbers
Example: 2.3 + 3.4+ 3.3 = 9
But output shows: 9.000000 + EEE or something like that.
So- how to convert, to be only 9, not this REAL variable.
To actually convert:
var
i: integer;
...
i := round(floatVar);
To output only the integer part:
writeln(floatVar:9:0);
Let's consider this quite simpler equation:
3.5 + 2.5
What do you expect? 6, right? Let's try this code
write(3.5 + 2.5);
Unfortunately, it's a floating-point number, so it would produce a number represented in a scientific way:
6.00000000000E+00
or 6.0000000000 x 100, or 6 x 10o. Whatever, you only care about 6, who need this weird useless long number? So the idea is to cut off the decimal part and output to the console only the integer part, which can be done with this line of code:
write(3.5 + 2.5 : 0 : 0);
Ok, now it outputs a beautiful number as expected
6
Seems like the problem is solved, but you say that:
I'm summing some numbers
Example: 2.3 + 3.4+ 3.3 = 9
Ohh so that the evenly, beautiful integer is just randomly appeared? Here the problem comes, how do you expect this equation would output?
3.6 + 2.5
It should be 6.1, right? Let's try it with the worked line of code:
write(3.6 + 2.5 : 0 : 0);
And the output is...
6
Unexpected, right? So how about rounding to some decimal places, like 1?
write(3.5 + 2.5 : 0 : 1);
write(3.6 + 2.5 : 0 : 1);
Then, 3.5 + 2.5 = 6.0 and 3.6 + 2.5 = 6.1. But 6.0 may look quite long, so how to make it output 6 for 6.0 and 6.1 for 6.1?
Actually, you can't make the program auto-detect if a real variable contains an integer value because the way a real var is stored is completely different from an integer var (how different they are, please contact Google; but you can do it manually by making a function to do the job).
So my solution is, to be easy, making the output rounded to some decimal places, and that's it.
For purpose of showing pretty output on the screen you can use something like this:
Writeln(result:0:2);
Result on screen would be this:
9.00
What this means someone would ask? Well first number 0 means how wide filed is. if you say it's 0 then Pascal writes it at the very left side of screen. If you said writeln(result:5:2) result would be:
9.00
In other words i would print form the right side and leave 5 chars to do so.
Second number 2, in this example means you want that result printed with 2 decimal places. You can place it only if you want to print on screen value that is real, single, double, extended and so on.You can round to any number of decimals, and if you do writeln(result:0:0) you would get ouput:
9
If you are printing integer and want to have some length of field, lets sat 5 you would do: writeln(int:5). If you added :2 to the end you would get compile time error.
This all also works for something like this: writeln(5/3.5+sqrt(3):0:3),
You should know that this does not round variable itself but just formats output. This is also legal:
program test;
var
a:real;
n,m:integer;
begin
readln(a,m,n);
writeln(a:m:n);
end.
What i did here is i asked user if on how many decimals and with what length of field he wants to write entered number a. This can be useful so i'm pointing it out. Thank you for reading. I hope i helped
You can convert to string, get the int part, e convert to int number!
Or Float to Str than Str to Int:
nPage := StrToInt(FloatToStr(Int(nReg / nTPages))) + 1;