I am storing a value (represented as a string originally) like this - 12345678901234.12345678912 - in a double variable. After storing, it is represented in an exponential format (with an e). How do i convert this exponential representation to the original(string) representation?
Dim s as string = "1234567891234567.123456789"
Dim d as Double
Double.TryParse(s, d)
Console.WriteLine(d) 'Prints 1.23456789123457E+15
Using Decimal solves the problem but why cant Double do it?
Your string contains 25 significant digits. double simply doesn't retain that amount of information. Even decimal can barely hold that much (28/29 digits). From the docs for System.Double:
By default, a Double value contains 15 decimal digits of precision, although a maximum of 17 digits is maintained internally.
You should read my articles on binary floating point and decimal floating point for more information - they come at the topic from a C# point of view, but you're obviously using the same types from VB.
In your particular case, the exact double value closest to 1234567891234567.123456789 is just 1234567891234567 - you're losing all the information after the decimal point.
Related
Executing the following statement results in Access SQL:
CLNG((CCUR(1.225)/1)*100) = 123
The Conversion Goes, Decimal > Currency > Double > Double > Long
If I remove the CCUR conversion function:
CLNG(((1.225)/1)*100) = 122
The Conversion here goes , Decimal > Double > Double > Long
What is the difference between these two?
This extends to being different between Code And Access SQL
In Access SQL
clng((CCUR(1.015)/1)*100)/100 = 1.01 (Wrong Rounding)
In Access VBA
clng((CCUR(1.015)/1)*100)/100 = 1.02 (Appropriate Rounding Here)
Microsoft explain that the CLng function uses Banker's Rounding, here.
When the fractional part is exactly 0.5, CInt and CLng always round it to the nearest even number. For example, 0.5 rounds to 0, and 1.5 rounds to 2. CInt and CLng differ from the Fix and Int functions, which truncate, rather than round, the fractional part of a number. Also, Fix and Int always return a value of the same type as is passed in.
Looking at a similar question and the subsequent answer HERE, it explains that there are changes to the bit calculation behind the scenes, based on how it is calculated, but I'm not sure how the data type effects it.
What am I missing, and why is it calculating this way? How could I reproduce this behavior predictably in SQL Server?
EDIT
After some digging I believe that this is truly the result of a rounding point issue. In SQL server it will round floats to the nearest whole number if it is outside of the 15 digit max of precision. Access seems to hold more somehow, even though a Double is equivalent to a Float(53) in TSQL.
The difference in results is a combination of two different issues: Jet/ACE vs VBA expression evaluation and binary floating point representation of decimal numbers.
The first is that the Jet/ACE expression engine implicitly converts fractional numbers to Decimal while VBA converts them to Double. This can be easily demonstrated (note the Eval() function evaluates an expression using the Jet/ACE db engine):
?Typename(1.015), eval("typename(1.015)")
Double Decimal
The second issue is that of floating point arithmetic. This is somewhat more difficult to demonstrate because VBA always rounds its output, but the issue is more obvious using another language (Python, in this case):
>>> from decimal import Decimal
>>> Decimal(1.015)
Decimal('1.0149999999999999023003738329862244427204132080078125')
The Double type in VBA uses floating-point arithmetic, while the Decimal type uses integer arithmetic (it stores the position of the decimal point behind the scenes).
The upshot to this is that Banker's rounding or traditional rounding is a red herring. The determining factor is whether the binary floating point representation of the number is slightly greater or less than its decimal representation.
To see how this works in your original question see the following VBA:
?Eval("typename((CCUR(1.225)/1))"), Eval("typename(((1.225)/1))")
Double Decimal
?Eval("typename(CCUR(1.225))"), Eval("typename(1.225)")
Currency Decimal
And Python:
>>> Decimal(1.225)
Decimal('1.225000000000000088817841970012523233890533447265625')
I should also point out that your assumption of the conversion to Double in your second example is incorrect. The data type remains Decimal until the final conversion to Long. The difference between the first two functions is that multiplying a Decimal by a Currency type in Jet/ACE results in a Double. This seems like somewhat odd behavior to me, but the code bears it out:
?eval("TypeName(1.225)"), eval("TypeName(1.225)")
Decimal Decimal
?eval("TypeName(CCUR(1.225))"), eval("TypeName((1.225))")
Currency Decimal
?eval("TypeName(CCUR(1.225)/1)"), eval("TypeName((1.225)/1)")
Double Decimal
?eval("TypeName((CCUR(1.225)/1)*100)"), eval("TypeName(((1.225)/1)*100)")
Double Decimal
?eval("TypeName(CLNG((CCUR(1.225)/1)*100))"), eval("TypeName(CLNG(((1.225)/1)*100))")
Long Long
So the conversion in the two cases is actually:
Decimal > Currency > Double > Double > Long (as you correctly assumed); and
Decimal > Decimal > Decimal > Decimal > Long (correcting your initial assumption).
To answer your question in the comment below, Eval() uses the same expression engine as Jet/ACE, so it is functionally equivalent to entering the same formula in an Access query. For further proof, I present the following:
SELECT
TypeName(1.225) as A1,
TypeName(CCUR(1.225)) as A2,
TypeName(CCUR(1.225)/1) as A3,
TypeName((CCUR(1.225)/1)*100) as A4,
TypeName(CLNG((CCUR(1.225)/1)*100)) as A5
SELECT
TypeName(1.225) as B1,
TypeName((1.225)) as B2,
TypeName((1.225)/1) as B3,
TypeName(((1.225)/1)*100) as B4,
TypeName(CLNG(((1.225)/1)*100)) as B5
I'm using VB.NET, writing a winforms application where I'm trying to convert from a denary real number to a signed floating point binary number, as a string representation. For example, 9.125 would become "0100100100000100" (the first ten digits are the significand and the last six digits the exponent.).
I can write a function for this if I have to, but I'd rather not waste time if there's a built-in functionality available. I know there's some ToString overload or something that works on Integers, but I haven't been able to find anything that works on Doubles.
I am trying to get the integer part of a number after dividing two variables.
ie, get 3 if the value is 3.75
displaycount and itemcount are both integer variables.
Dim cntr As Integer
cntr = Math.Floor(Math.Abs(itemCount / displaycount))
That code produces a blue squiggly in VS2012 with the comment that "runtime errors may occur when converting Double to Integer" BUT Math.Floor is supposed to take a decimal or double and return an integer.
"Math.Floor is supposed to take a decimal or double and return an integer." No, it isn't. It returns a value of the same type as its argument. See the documentation, e.g. Math.Floor Method (Double).
I would have expected VS to suggest a fix of adding CInt() around the RHS of the assignment; did that not appear for you?
If you need an Integer as result, consider using either the CInt, Int or the Fix functions.
CInt rounds to the nearest integer using the bankers's rounding (n.5 rounds towards the closest even number).
Int removes the fractional parts. Negative numbers are truncated towards smaller numbers
Int(-8.4) = -9.
Fix removes the fractional parts. Negative numbers are truncated towards greater numbers
Fix(-8.4) = -8.
See Conversion.Int Method and Type Conversion Functions (Visual Basic).
When two variables are declared as integer type and you perform
14/4, you get 4, but when you use integer division, 14\4, you get 3.
I thought when you use integer division it rounds to the closest even number. So 14\4 = 3.5 (4 is the closest even number) should be 4 instead,
right?
In VB.NET, the / operator is defined to return a floating-point result. It converts the variables to double before performing the division.
This is not the case in the integer division \ where the division is performed without the remainder if the quotient is a decimal (decimals are ignored). For example if the quotient is 3.x, then x is ignored
When you cast a floating point number to an integer in VB.NET, the value is rounded to the nearest even number. Apparently rounding a number when converting it to an integer is a behavior that stretches back to the days of the BASIC language.
However, when performing integer division (with the \ operator), the fractional part is simply discarded, no matter what the fractional part is. This is why you get the behavior that you are seeing.
Please forgive me, I haven't used this site very much! I am working in Visual Studio with Visual Basic. I finished programming my project with Option Strict Off, then when I turned Option Strict on, I was alerted that this code was wrong:
Const TAX_Decimal As Decimal = 0.07
The explanation was that "Option Strict On disallows implicit conversions from 'Double' to 'Decimal'"
But I thought I had declared it as a decimal! It made me change it to:
Const TAX_Decimal As Decimal = CDec(0.07)
The only thing I did with this constant was multiply it by a decimal and saved it to a variable declared as a decimal!
Can someone tell me why this is happening?
Double is 8 bytes and Decimal is 16 bytes. Option Strict prevents from automatic type conversion. By default if you write a number with decimals in VB.NET it is considered as double and not decimal. For saying decimal you have to use some character to specify (I thing for decimal is m) so if you declare
Const VAR as decimal = 0.07m
then you wont require casting.
When the compiler sees a numeric literal, it selects a type based upon the size of the number, punctuation marks, and suffix (if any), and then translates the the sequence of characters in it to that type; all of this is done without regard for what the compiler is going to do with the number. Once this is done, the compiler will only allow the number to be used as its own type, explicitly cast to another type, or in the two cases defined below implicitly converted to another type.
If the number is interpreted as any integer type (int, long, etc.) the compiler will allow it to be used to initialize any integer type in which the number is representable, as well as any binary or decimal floating-point type, without regard for whether or not the number can be represented precisely in that type.
If the number is type Single [denoted by an f suffix], the compiler will allow it to be used to initialize a Double, without regard for whether the resulting Double will accurately represent the literal with which the Single was initialized.
Numeric literals of type Double [including a decimal point, but with no suffix] or Decimal [a "D" suffix not followed immediately by a plus or minus] cannot be used to initialize a variable of any other, even if the number would be representable precisely in the target type, or the result would be the target type's best representation of the numeric literal in question.
Note that conversions between type Decimal and the other floating-point types (double and float) should be avoided whenever possible, since the conversion methods are not very accurate. While there are many double values for which no exact Decimal representation exists, there is a wide numeric range in which Decimal values are more tightly packed than double values. One might expect that converting a double would choose the closest Decimal value, or at least one of the Decimal values which is between that number and the next higher or lower double value, but the normal conversion methods do not always do so. In some cases the result may be off by a significant margin.
If you ever find yourself having to convert Double to Decimal, you're probably doing something wrong. While there are some operations which are available on Double that are not available on Decimal, the act of converting between the two types means whatever Decimal result you end up with is apt to be less precise than if all computations had been done in Double`.