I'm making a simple calculator code, and when I do the division I want it to show not just the whole number but the decimal number.
This is my division code:
get1.Text = Int(mygive.Text \ rate.Text)
I've also tried:
get1.Text = Int(mygive.Text / rate.Text)
I want it to show numbers like this: 2060.0891
Thanks in advance!
You need to convert the numbers to Double prior to doing the division. CDbl is an appropriate Type Conversion Function for this:
Dim answer = CDbl(mygive.Text) / CDbl(rate.Text)
get1.Text = answer.ToString()
Your Int is converting everything to whole numbers. Try converting to double.
It's showing only the whole part of the number because you're declaring an Integer, which is insufficient, as it can only store whole numbers.
You need to use the Double data type.
Also, you're attempting to apply arithmetic operations on Strings, which is invalid.
You need to convert those strings to be of the type Double before you can do / on them.
Use converting to Double/Decimal
with VB help function:
get1.Text = CDec(mygive.Text / rate.Text).ToString()
with .NET function
get1.Text = Convert.ToDecimal(mygive.Text / rate.Text).ToString()
Related
The below line of code is returning as a 1 instead of a 2, for reasons I can't comprehend.
console.write(Format((87.20 \ 43.60)))
Surely this should return the result of 2 but I've checked in another environment and it returns a can anyone tell me why?
I have tried putting the code into a second environment but the result was the same I don't understand why it is returning 1 instead of 2, can anyone enlighten me?
Thanks for the help but found the answer.
Decimals are converted to Long before Integer division and due to this are subject to banker's rounding, multiplying both numbers by a 100 before the operation resolves the issue.
Source of information:
Learn Microsoft - VB.Net - \ Operator - Remarks
Thanks,
While I am glad you resolved your own question, I did want to provide an alternative.
When you use the integer division operator, you do not have control as to how the rounding should occur. Whereas if you do a floating-point division operator you can keep the precision and then use one of the built-in Math class methods (documentation) to specify how the number should round.
Take a look at this example:
Dim result = 87.20 / 43.60
Dim roundUp = Math.Ceiling(result)
Dim roundDown = Math.Floor(result)
Dim bankersRounding = Math.Round(result)
Fiddle: https://dotnetfiddle.net/LZEMXV
Because in your example you are using Console.Write (which treats the data as a String) you do not need to cast the value to an Integer data type. However, if you needed the value as an Integer, any one of those variables outside result can be safely converted to an Integer using the Parse method (documentation).
I'm trying to get a hash of a decimal value and convert it to an integer. But the query results in the following error:
Numeric value 'b902cc4550838229a710bfec4c38cbc7eb11082367a409df9135e7f007a96bda' is not recognized
SELECT (CAST(sha2(TO_VARCHAR(ABS(12.5)), 256) AS INTEGER) % 100) AS temp_value
What is the correct way to convert a hash string to an integer in Snowflake?
I can not use any user defined functions. And have to go with Snowflake native functions.
The hash value contains alphabetic character so it will throw an error
SELECT --(CAST(
sha2(
TO_VARCHAR(
ABS(12.5)), 256)-- AS INTEGER) % 100)
AS temp_value;
You need to convert the hex value from the hash encoding to be int.
I've not been able to find a function built into Snowflake that does this, but if you have a look in the following link, it will explain how to create a javascript function to do the conversion for you:
https://snowflakecommunity.force.com/s/article/faq-does-snowflake-have-a-hex-to-int-type-function
If you use the function in the link, then your code becomes something like this:
SELECT (CAST(js_hextoint(sha2(TO_VARCHAR(ABS(12.5)), 256)) AS INTEGER) % 100) AS temp_value
I've not been able to test the above code I'm afraid, so there may be a bracket in the wrong place...
You have a 56 digit hexadecimal number. It's not going to fit into the maximum numeric precision of 38. You could use a floating point number, but that will lose precision.
create or replace function CONV(VALUE_IN string, OLD_BASE float, NEW_BASE float)
returns string
language javascript
as
$$
// Usage note: Loses precision for very large inputs
return parseInt(VALUE_IN, Math.floor(OLD_BASE).toString(Math.floor(NEW_BASE)));
$$;
select conv('b902cc4550838229a710bfec4c38cbc7eb11082367a409df9135e7f007a96bda', 16, 10);
--Returns 8.368282050700398e+76
For hex to integer, check Does snowflake have a ‘hex’ to ‘int’ type native function?. My guess is that most people checking this question (1k views) are looking for that.
But this specific question wants to convert a sha2 digest to integer for comparison purposes. My advice for that specific question is "don't".
That's because the hex string in the question represent the integer 83682820507003986697271120393377917644380327831689630185856829040117055843290, which is too much to handle even by Java's BigInteger.
Instead, just compare strings/binary to check if the values match or not.
What is the difference between:
Dim a As Integer = CInt(2.2)
and
Dim a As Integer = Math.Round(2.2)
?
CInt returns an integer but will round the .5 to nearest even number so:
2 = CInt(2.5)
4 = CInt(3.5)
Are both true, which might not be what you want.
Math.Round can be told to round away from zero. But returns a double, so we still need to cast it
3 = CInt(Math.Round(2.5, MidpointRounding.AwayFromZero))
There is bigger differences in CInt(), Int() and Round()... and others.
Round has parameters of rounding, so it is flexible and user friendly. But it do not change variable type. No "type conversion".
Meanwhile CInt() is a bit cryptic as it rounds too. And it is doing "Type conversion" to integer.
2 = Int(2.555), 3 = CInt(2.555)
2 = Int(2.5), 2 = CInt(2.5)
Some documentation states:
When the fractional part of expression is exactly .5, CInt always rounds it to the nearest even number. For example, .5 rounds to 0, and 1.5 rounds to 2.
But I do not like that "exact 0.5", in real word it is "0.5000001"
So, doing integer math (like calculating bitmaps address Hi and Lo bytes) do not use CInt(). Use old school INT(). Until you get to negative numbers... see the fix() function.
If there is no need to convert type, use floor().
I think all this chaos of number conversion is for some sort of compatibility with some ancient software.
The difference between those two functions is that they do totally different things:
CInt converts to an Integer type
Math.Round rounds the value to the nearest Integer
Math.Round in this instance will get you 2.0, as specified by the MSDN documentation. You are also using the function incorrectly, see the MSDN link above.
Both will raise an Exception if conversion fails, you can use Try..Catch for this.
Side note: You're new to VB.NET, but you might want to try switching to C#. I find that it is a hybrid of VB.NET & C++ and it will be far easier for you to work with than VB.NET.
Converting a floating-point number to an integer using either CInt or CType will cause the value of that number to be rounded. The Int function and Math.Floor may be used to convert a floating-point number to a whole number, rounding toward negative infinity, but both functions return floating-point values which cannot be implicitly used as Integer values without a cast.
Is there a concise and idiomatic alternative to IntVar = CInt(Int(FloatingPointVar));? Pascal included Round and Trunc functions which returned Integer; is there some equivalent in either the VB.NET language or in the .NET framework?
A similar question, CInt does not round Double value consistently - how can I remove the fractional part? was asked in 2011, but it simply asked if there was a way to convert a floating-point number to an integer; the answers suggested a two-step process, but it didn't go into any depth about what does or does not exist in the framework. I would find it hard to believe that the Framework wouldn't have something analogous to the Pascal Trunc function, given that such a thing will frequently be needed when performing graphical operations using floating-point operands [such operations need to be rendered as discrete pixels, and should be rounded in such a way that round(x)-1 = round(x-1) for all x that fit within the range of +/- (2^31-1); even if such operations are rounded, they should use Floor(x+0.5), rather than round-to-nearest-even, so as to ensure the above property]
Incidentally, in C# a typecast from Double to Int using (type)expr notation uses round-to-zero semantics; the fact that this differs from the VB.NET behavior suggests that one or both languages is using its own conversion routines rather an explicit conversion operator included in the Framework. It would seem likely that the Framework should define a conversion operator? Does such an operator exist within the framework? What does it do? Is there a way to invoke it from C# and/or VB.NET?
After some searching, it seems that VB has no clean way of accomplishing that, short of writing an extension method.
The C# (int) cast translates directly into conv.i4 in IL. VB has no such operators, and no framework function seems to provide an alternative.
Usenet had an interesting discussion about this back in 2005 – of course a lot has changed since then but I think this still holds.
You can use the Math.Truncate method.
Calculates the integral part of a specified double-precision floating-point number.
For example:
Dim a As double = 1.6666666
Dim b As Integer = Math.Truncate(a) ' b = 1
I know this is an old case but I saw no one suggest the Math.Round() function.
Yes Math.Round takes a double and returns a double. However it returns a number that has been rounded to a whole number. It should easily and concisely convert to an integer using cInt. Would that suffice?
cInt(math.round(10000.54564)) ' = 10001
cInt(math.round(10000.49564)) ' = 10000
You may need extract the Int part of a float number:
float num = 12.234;
string toint = "" + num;
string auxil = toint.Split('.');
int newnum = Int.Parse(auxil[0]);
I have a value of: "2.54334881002458E-37" and i keep getting "overflow" exception when i'm using a double.
what should i use to make this work?
Thank you
code snippet:
Dim curries, act, cat As Double
For Each dataRow As DataRow In dt.Rows
curries = dataRow("Activity")
getting the error when i try to assign Activity to curries.
but "activity" is a string in the database....
Double is already 64 bits worth of floating point number.
Can you post code where you are getting this overflow?
Decimal might be worth a shot, but you have to post code so that we can understand the issues you are encountering.
Based on your edit in your post, why are you storing numbers as strings in your database? That is a definite no no...unless you are not doing any sort of arithmaetic operation only then can you store them as varchar / string.
Give us a sample of what the data looks like...I think your issue stems from not converting the string to a decimal, if activity is a string convert it using DirectCast or CType (cast the value):
curries = CType(datarow("Activity"), Double)
Change your unit of measure, so that you're not working in 10^-37 of whatever it is you're dealing with. This problem just screams "I'm not solving this in the appropriate domain."
According to MSDN, Double should have no problem at all with your number:
Range:
-1.79769313486231570E+308 through -4.94065645841246544E-324 † for negative values;
4.94065645841246544E-324 through 1.79769313486231570E+308 † for positive values
and if the need to have large numbers and precise I suggest the use of decimal numbers, more information here.
http://msdn.microsoft.com/en-us/library/system.decimal.aspx
Bye