I know about the Math.Round and Math.Ceiling methods, but they return a Double and a Decimal. Does VB.NET have any built-in functions which always round a floating point number up, not down, with a return type of Integer? I know there's CInt, but this can round down if it's below 6.5.
From the comments I understood you wanted to round 6.1 to 7.
Just add 1 and truncate.
If it looks awkward, create a method for it.
Correction:
Unless the number already is truncated.
Addendum:
Note here that doing == with floats is not without problem; you should always have some sort of precision when trying == with floats.
Now when you have decided on this precision - then you can rewrite your code to add 0.999999 (according to precision) and the first add-0.9999-and-truncat works.
Note again that adding 0.9999 is does not really mean you add 0.9999 with our "normal" float.
So if you really want to add 0.9999 you have to work with BND and/or some monetary arithmetics.
Which you "always" should do when calculating money (or any exact decimal stuff)
Related
I can find a workaround, but it is really annoying and I may certainly be missing something. Redshift's ROUND function doesn't round to the number of decimals specified.
For example,
select round(cast(176 as float)/cast(492 as float),4) as result;
Above select statement will return 0.35769999999999996.
However, this statement:
select round(cast(229 as float)/cast(491 as float),4) as result;
... will return 0.4664.
Why? I can work around this, but seems like it should work and return only four decimal places.
If your issues is all those 9999s, then the issue is floating point representation. Convert to a decimal to get fixed-point precision:
select round(cast(176 as float)/cast(492 as float), 4)::decimal(10, 4) as result;
Elaborating more on Gordon's answer -
So you’ve written some absurdly simple code, say for example:
0.1 + 0.2
and got a really unexpected result:
0.30000000000000004
Because internally, computers use a format (binary floating-point) that cannot accurately represent a number like 0.1, 0.2 or 0.3 at all.
When the code is compiled or interpreted, your “0.1” is already rounded to the nearest number in that format, which results in a small rounding error even before the calculation happens.
What can I do to avoid this problem?
That depends on what kind of calculations you’re doing.
If you really need your results to add up exactly, especially when you work with money: use a decimal datatype.
If you just don’t want to see all those extra decimal places: simply format your result rounded to a fixed number of decimal places when displaying it.
Shamelessly stolen from : Floating Point
try multiplying by 10 to the power of your desired places after the decimal point, rounding, and then dividing it out again:
-- exclude decimal point inside ROUND(), include outside ROUND()
SELECT ROUND(10000 * 176 / 492) / 10000.0
which will return the expected 0.3577.
I am trying to work with decimals
If "3.04" < "12.4" Then
finalPrice = "perfect"
Else
finalPrice = "too big"
End If
So 3.04 is not bigger than 12.4 right? When I run this it's thinking 3.04 IS bigger than 12.4. Why is it doing that? It should return perfect instead of returning too big which is what it's currently doing.
Is it a decimal issue?
You cannot compare strings in that way (which is what it is, being encased in quotes). As this is a string comparison, lexicographically, "3" (the leftmost character) is higher than "1". Try parsing the numbers into a floating point number, then your comparison will work.
Additionally, remember, parse safely! If the parse fails, prepare to have some defensive coding around it. If anything, avoid this floating point number being entered in string form at all if possible.
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]);
Can any one please help me how to get float value as it is from text box
for Ex: I have entered 40.7
rateField=[[rateField text] floatValue];
I am getting rateField value as 40.7000008 but I want 40.7 only.
please help me.
thanks in advance
Thanks Every body,
I tried all the possibilities but I am not able to get what I want. I am not looking to print the value to convert into string.I want to use that value for computation. If i use Number Formatter again when i am converting from number to float it is giving same problem.So i want float value only but it should be whatever i have given in the text box it should not be padded with any values.This is my requirement.Please help me.
thanks®ards Balu
Thanks Every body,
I tried all the possibilities but I am not able to get what I want. I am not looking to print the value to convert into string.I want to use that value for computation. If i use Number Formatter again when i am converting from number to float it is giving same problem.So i want float value only but it should be whatever i have given in the text box it should not be padded with any values.This is my requirement.Please help me.
thanks®ards
Balu
This is ok. There is not guaranteed that you will get 40.7 if you will use even double.
If you want to output 40.7 you can use %.1f or NSNumberFormatter
Try using a double instead. Usually solves that issue. Has to do with the storage precision.
double dbl = [rateField.text doubleValue];
When using floating point numbers, these things can happen because of the way the numbers are stored in binary format in the computers memory.
It's similar to the way 1/3 = 0.33333333333333... in decimal numbers.
The best way to deal with this is to use number formatters in the textbox that displays the value.
You are already resolved float value.
Floating point numbers have limited precision. Although it depends on
the system, float relative error due to rounding will be around 1.1e-8
Non elementary arithmetic operations may give larger errors, and, of
course, error progragation must be considered when several operations
are compounded.
Additionally, rational numbers that are exactly representable as
floating point numbers in base 10, like 0.1 or 0.7, do not have an
exact representation as floating point numbers in base 2, which is
used internally, no matter the size of the mantissa. Hence, they
cannot be converted into their internal binary counterparts without a
small loss of precision. This can lead to confusing results: for
example, floor((0.1+0.7)*10) will usually return 7 instead of the
expected 8, since the internal representation will be something like
7.9999999999999991118....
So if you're using those numbers for output, you should use some rounding mechanism, even for double values.
I here is the problem:
I have a NSString that contain "1.7" (for example) and I have to get the float number = 1.7
I' ve tried with [mystring floatValue] but the result is 1.700000000004576
If I try with "1.74" the result is 1.74000000000000000067484
how can I fix it?
thank you!
You are correctly converting the string into a float. The problem is that floating point numbers cannot represent all real numbers exactly. A direct assignment:
float x = 1.7;
will still have a precision error. That's just how floating point numbers are.
The workaround depends on your needs. Some examples: If you need more precision for mathematical calculations, you can use doubles. If you're trying to generate output for the user, you can format the output so it limits the number of digits shown after the decimal point. If you're dealing with money, you could convert floating point dollar amounts into integer numbers of cents and perform all calculations using integers, only showing a decimal point on output to the user.
Floats need to be able to represent infinitely many real numbers, but a float contains a finite number of bits, so floats are approximations.
See this article for more.
You can trim the answer by using the formatting below.
The .1 will set the result to one decimal place.
NSLog(#"mystring = %.1f ",[mystring floatValue]);
Solved!
I used sqlite3_bind_text even if my database attribute is FLOAT value and I used mystring instead of myfloat and it work fine! Thank you!