In the debug window, when I input this command:
po 1912/10.0
The output is 191.19999999999999.
What I really want to get back is 191.2.
Why is this happening, and how can I convert an int into a double with precision?
From What Every Programmer Should Know About Floating-Point Arithmetic:
Why don’t my numbers, like 0.1 + 0.2 add up to a nice round 0.3, and instead I get a weird result like 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.
This is why programmers say you should only ever store money as an integer. For example int cents = 1995; rather than float dollars = 19.95.
If your app doesn't need to be 100% precise (for example, if you're calculating screen coordinates or translucency or a color) just format your float rounded to 1 or 2 decimal places:
double someValue = 1912/10.0;
NSLog(#"2 decimals: %.2f", someValue);
NSLog(#"0 decimals: %.0f", someValue);
This code will output:
2 decimals: 191.20
0 decimals: 191
That's normal for a floating point number. Double is obviously just an extended precision floating point number. If you want to keep the pristine decimal digits, then don't allow any float/double conversion. Instead store the result as a scaled integer (in your case 1912) and place the decimal manually.
Let me try to explain this another way. When you express a number with a fractional part with a float or double, precision is most often lost. There's no way around that. If you store 1912 as a float and store 10 as a float then divide the first stored value by the second, the value will NEVER be 191.2. That's just the way floating point numbers work. If you look at the number in a debugger you'll see something like 191.19999999999999 as you describe. This, in itself, is an approximation as the value should be 191.19999999999999... but of course you can't even type all the digits in the decimal value of that stored result as the number of digits approaches infinity.
If you're going to use floating point, that's what you'll get. No way around it.
If you really want to get 191.2, then you can't use floating point, at least without doing rounding. Instead, you need to normalize the numbers by just storing the value as 1912 and printing the value with a decimal point to the left of the 2.
There's another brief online description at http://floating-point-gui.de/basic/
Related
In Kotlin 123.456 is a valid Double value, however, 123.456F.toDouble() results in 123.45600128173828 - presumably just the way precision is handled between the two.
I want to be able to convert freely between the two, specifically for cases like this:
123.456F -> 123.456 // Float to Double
123.456 -> 123.456F // Double to Float
How can I convert a float to a double in cases like this, and maintain precision?
It's a big ugly, but you could convert your Float to a String and back out to a Double:
val myDouble: Double = 123.456f.toString().toDouble()
// 123.456d
You could always encapsulate this in an extension function:
fun Float.toExactDouble(): Double =
this.toString().toDouble()
val myDouble = 123.456f.toExactDouble()
In Kotlin 123.456 is a valid Double value
Actually, that's not quite true. There's a Double value very close to 123.456, but it's not exactly 123.456. What you're seeing is the consequences of that.
So you can't maintain precision, because you don't have that precision to start with!
Short answer:
If you need exact values, don't use floating-point!
(In particular: Never store money values in floating-point! See for example this question.)
The best alternative is usually BigDecimal which can store and calculate decimal fractions to an arbitrary precision. They're less efficient, but Kotlin's operator overloading makes them painless to use (unlike Java!).
Or if you're not going to be doing any calculations, you could store them as Strings.
Or if you'll only need a certain number of decimal places, you could scale them all up to Ints (or Longs).
Technical explanation:
Floats and Doubles use binary floating-point; they store an integer, and an integer power of 2 to multiple or divide it by. (For example, 3/4 would be stored as 3*2⁻².) This means they can store a wide range of binary fractions exactly.
However, just as you can't store 1/3 as a decimal fraction (it's 0.3333333333…, but any finite number of digits will only be an approximation), so you can't store 1/10 as a binary fraction (it's 0.000110011001100…). This means that a binary floating-point number can't store most decimal numbers exactly.
Instead, they store the nearest possible value to the number you want. And the routines which convert them to a String will try to undo that difference, by rounding appropriately. But that doesn't always give the result you expect.
Floating-point numbers are great when you need a huge range of values (e.g. in scientific and technical use), but don't care about storing them exactly.
Right now I have a line of code like this:
float x = (([self.machine micSensitivity] - 0.0075f) / 0.00025f);
Where [self.machine micSensitivity] is a float containing the value 0.010000
So,
0.01 - 0.0075 = 0.0025
0.0025 / 0.00025 = 10.0
But in this case, it keeps returning 9.999999
I'm assuming there's some kind of rounding error but I can't seem to find a clean way of fixing it. micSensitivity is incremented/decremented by 0.00025 and that formula is meant to return a clean integer value for the user to reference so I'd rather get the programming right than just adding 0.000000000001.
Thanks.
that formula is meant to return a clean integer value for the user to reference
If that is really important to you, then why do you not multiply all the numbers in this story by 10000, coerce to int, and do integer arithmetic?
Or, if you know that the answer is arbitrarily close to an integer, round to that integer and present it.
Floating-point arithmetic is binary, not decimal. It will almost always give rounding errors. You need to take that into account. "float" has about six digit precision. "double" has about 15 digits precision. You throw away nine digits precision for no reason.
Now think: What do you want to display? What do you want to display if the result of your calculation is 9.999999999? What would you want to display if the result is 9.538105712?
None of the numbers in your question, except 10.0, can be exactly represented in a float or a double on iOS. If you want to do float math with those numbers, you will have rounding errors.
You can round your result to the nearest integer easily enough:
float x = rintf((self.machine.micSensitivity - 0.0075f) / 0.00025f);
Or you can just multiply all your numbers, including the allowed values of micSensitivity, by 4000 (which is 1/0.00025), and thus work entirely with integers.
Or you can change the allowed values of micSensitivity so that its increment is a fraction whose denominator is a power of 2. For example, if you use an increment of 0.000244140625 (which is 2-12), and change 0.0075 to 0.00732421875 (which is 30 * 2-12), you should get exact results, as long as your micSensitivity is within the range ±4096 (since 4096 is 212 and a float has 24 bits of significand).
The code you have posted is correct and functioning properly. This is a known side effect of using floating point arithmetic. See the wiki on floating point accuracy problems for a dull explanation as to why.
There are several ways to work around the problem depending on what you need to use the number for.
If you need to compare two floats, then most everything works OK: less than and greater than do what you would expect. The only trouble is testing if two floats are equal.
// If x and y are within a very small number from each other then they are equal.
if (fabs(x - y) < verySmallNumber) { // verySmallNumber is usually called epsilon.
// x and y are equal (or at least close enough)
}
If you want to print a float, then you can specify a precision to round to.
// Get a string of the x rounded to five digits of precision.
NSString *xAsAString = [NSString stringWithFormat:#"%.5f", x];
9.999999 is equal 10. there is prove:
9.999999 = x then 10x = 99.999999 then 10x-x = 9x = 90 then x = 10
Here is my question :
If we have the following value
0.59144706948010461
and we try to convert it to Single we receive the next value:
0.591447055
As you can see this is not that we should receive. Could you please explain how does this value get created and how can I avoid this situation?
Thank you!
As you can see this is not that we should receive.
Why not? I strongly suspect that's the closest Single value to the Double you've given.
From the documentation for Single, having fixed the typo:
All floating-point numbers have a limited number of significant digits, which also determines how accurately a floating-point value approximates a real number. A Single value has up to 7 decimal digits of precision, although a maximum of 9 digits is maintained internally.
Your Double value is 0.5914471 when limited to 7 significant digits - and so is the Single value you're getting. Your original Double value isn't exactly 0.59144706948010461 either... the exact values of the Double and Single values are:
Double: 0.5914470694801046146693579430575482547283172607421875
Single: 0.591447055339813232421875
It's important that you understand a bit about how binary floating point works - see my articles on binary floating point and decimal floating point for more background.
When converting from double to float you're also rounding. The result should be the single-precision number that is closest to the number you are rounding.
That is exactly what you're getting here.
Floating-point numbers between 0.5 and 1 are of the form n / 2^24 where n is between 2^23 and 2^24.
0.59144706948010461... = 9922835.23723472274456576... / 2^24
so the closest single-precision floating-point number is
9922835 / 2^24 = 0.5914470553...
I have 2 buttons that each have a tag number that I pass into this string in which I am just trying to type in either 1,1,1,1,1,1,1,1,1 or 2,2,2,2,2,2,2 or shoot - even, 1,2,2,1,1,1.
Everything works fine until the 8th or 9th time of pressing the button "1" the label says, 111111112. Then if I press the 1 again the label says, 111111168.
Maybe I am going about this totally wrong? Made sense in my head - but now I am just confused. Any help would be amazing, thank you!
-(IBAction)buttonDigitPressed:(id)sender {
currentNumber=currentNumber * 10 + (float)[sender tag];
NSLog(#"currentNumber: %.f", currentNumber);
phoneNumberLabel.text = [NSString stringWithFormat:#"%.f",currentNumber];
}
This image shows me hitting the 1 a bunch of times.. you'd think it would just keep showing 1's all the way across, no?
If this is a string operation, you should not do it using numbers. Possible reasons of the error: running out of range (because float is not big enough), loss of precision (because of the nature of float), etc. What you should do instead is
phoneNumberLabel.text = [phoneNumberLabel.text stringByAppendingFormat:#"%d", [sender tag]];
(Single precision) floating point numbers use 23 bits for the mantissa, therefore the largest integer that can be represented exactly by a float is 2^24 = 16777216.
All larger integers can not be represented exactly by a float, therefore the calculation with numbers having 8 or more digits using float cannot be exact.
Double precision floating point numbers can represent numbers up to 2^53 = 9007199254740992 exactly.
A better solution might be to work with integer types (e.g. uint64_t), or with strings as suggested in H2CO3's answer.
I have NSSlider, and I rect to it's changes. But I want my action to work only if slider's float value is for example 2.230000 or 3.410000.
if (floatValue is y.xx0000) {
doSomething;
}
I mean I want to do some action only if my float has only 2 decimal places not equal to 0. How could I do it?
The only float values that have only two non-zero fractional digits are numbers of the form n.00, n.25, n.50, or n.75. All other values have more than two non-zero fractional digits. Your example, 3.41, for example, isn't really "3.41". Instead it's:
3.410000000000000142108547152020037174224853515625
and "2.23" is actually:
2.229999999999999982236431605997495353221893310546875
So what are you really trying to do?
I'm a bit late to this game, but I found this question while searching for something similar.
It sounds to me like what you're trying to do is to round the value that the slider is set to to 2 decimal places. That way the thing and/or calculation you're trying to configure with the slider, will always work as though it only works for exact 2 decimal values.
Alternatively, you could check how large the distance is between your float value, and the value rounded to 2 decimal places. And then put a threshold of for example 0.005 for when it's to far away.