I was converting Float => CGFloat and it gave me following result. Why It comes as "0.349999994039536" after conversion but works fine with Double => CGFloat?
let float: Float = 0.35
let cgFloat = CGFloat(float)
print(cgFloat)
// 0.349999994039536
let double: Double = 0.35
let cgFloat = CGFloat(double)
print(cgFloat)
// 0.35
Both converting “.35” to float and converting “.35” to double produce a value that differs from .35, because the floating-point formats use a binary base, so the exact mathematical value must be approximated using powers of two (negative powers of two in this case).
Because the float format uses fewer bits, its result is less precise and, in this case, less accurate. The float value is 0.3499999940395355224609375, and the double value is 0.34999999999999997779553950749686919152736663818359375.
I am not completely conversant with Swift, but I suspect the algorithm it is using to convert a CGFloat to decimal (with default options) is something like:
Produce a fixed number of decimal digits, with correct rounding from the actual value of the CGFloat to the number of digits, and then suppress any trailing zeroes. For example, if the exact mathematical value is 0.34999999999999997…, and the formatting uses 15 significant digits, the intermediate result is “0.350000000000000”, and then this is shorted to “0.35”.
The way this operates with float and double is:
When converted to double, .35 becomes 0.34999999999999997779553950749686919152736663818359375. When printed using the above methods, the result is “0.35”.
When converted to float, .35 becomes 0.3499999940395355224609375. When printed using the above method, the result is “0.349999994039536”.
Thus, both the float and double values differ from .35, but the formatting for printing does not use enough digits to show the deviation for the double value, while it does use enough digits to show the deviation for the float value.
Related
I've been writing a CAD type program for fun in JAVA. The other day I wrote some code to define a line which was tangent to 2 circles. I've been checking my numbers with a commercial CAD program and they have been fairly close. Usually to the 9th decimal point. My results really only need to be stored in an array to 7 decimal points. After successfully defining the line tangent to the 2 circles, I decided to test it and define a point which was the intersection of the line and one of the circles.
In one case I got the result I was looking for, in another case I got no intersection. After looking at a few of the calculations I realized I was getting a very very small variation of maybe 9 or ten decimal places. I'm thinking of rewriting the code using BigDecimal.
This is a small snippet of some of the code I need to rewrite. Once I started it became much more cumbersome than I was wanting to do. I'm thinking about just converting the results using BigDecimal and using the original code unless there is an easy way to convert the following code to a BigDecimal type of format.
private float[] offsetLine(float lnx1, float lny1, float lnz1, float lnx2, float lny2, float lnz2, String direction, float offset) {
double deltax = Math.abs(lnx2 - lnx1);
double deltay = Math.abs(lny2 - lny1);
double lineLength = Math.sqrt(deltax * deltax + deltay * deltay);
double stepx = (offset * deltay) / lineLength;
double stepy = (offset * deltax) / lineLength;
Ok I'll answer my own question. Here's some code I dug up. I could only round to 6 decimal places to get the rounding I wanted. Once I did my calculations in double values I called the subroutine roundDbl
double checkRadius1 = Math.sqrt(((cir1x - offsetpts[0])*(cir1x - offsetpts[0])) + ((cir1y - offsetpts[1]) * (cir1y - offsetpts[1])));
double checkRadiusRounded = roundDbl(checkRadius1, 6); //round to 6 decimal places
public static Float roundDbl(Double dblValue, int decimalPlace) {
String tempDblString = Double.toString(dblValue);
String tempDbl = new BigDecimal(tempDblString).setScale(decimalPlace, RoundingMode.HALF_UP).stripTrailingZeros().toPlainString();
return Float.valueOf(tempDbl);
}
I got an error when trying to multiply a number with a floating point in dart. Does anyone know why this happens?
void main() {
double x = 37.8;
int y = 100;
var z = x * y;
print(z);
// 3779.9999999999995
}
In other languages (C#/C++) I would have the result: 3780.0
This is completely expected because 37.8 (or rather, the 0.8 part) cannot be precisely encoded as a binary fraction in the IEEE754 standard so instead you get a close approximation that will include an error term in the LSBs of the fraction.
If you need numbers that are lossless (e.g. if you are handling monetary calculations) then check out the decimal package.
A simpler hack if your floating point number has sufficient bits allocated to the fraction to keep the erroroneous bits out of the way is to round off the number after your calculation to the number of decimal places that you care about.
You can use toStringAsFixed to control fraction digits.
void main() {
double x = 37.8;
int y = 100;
var z = x * y;
print(z.toStringAsFixed(1));
// 3780.0
}
I try to find the steps between a min and a max value with a given step-size, using swift 2.1.
So we have a min and a max value, both of type Double. The step-size is a Double too. If min is 0.0 and max 0.5 with steps of 0.1, the result is 6, obviously.
But if I start with -0.1 as the minimum value, the result is 6 too. But should be 7, agree?
Here is my Playground example:
let min:Double = -0.1
let max:Double = 0.5
let step:Double = 0.1
var steps: Int {
return Int((max - min) / step) + 1
}
print("steps: \(steps)") // returns "steps: 6", but should be 7
The result is 6.99999999 if we use a Double for the steps variable. But this loss of precision only occurs when our min value is negative.
Do you know a workaround? I just don't want to round() each time I calculate with Doubles.
When you use Int() it forces truncation of your number, which always rounds towards zero. So, 6.9999 becomes 6 rather than 7, because it's closer to zero. If you use round() first it should help:
var steps: Int {
return Int(round((max - min) / step) + 1.0)
}
That's always not a good idea to calculate integral steps based on floating point ranges, you'll always encounter issues, and you won't be able to do much.
Instead I recommend to build your logic on integral steps, and calculate double values based on integral values (not vice versa as you do). I.e. you don't calculate integral step based on range, but you set your integral number of steps and calculate your double step.
I am developing an app and want to round off values
i.e if the output is 4.8 I want to display 4.8
while if the output is 4.0 , I want to display 4
Also, it would be great if I could precisely round values : as in if value is 4.34 then round to 4.3 while if its 4.37 then round it to 4.4
One way to round floating point values is to just add 0.5 and then truncate the value.
double valueToRound = GetTheValueFromSomewhere();
double roundedValue = (double)((int)(valueToRound + 0.5));
This will round 1.4 down to 1.0 and 1.5 up to 2.0 for example. To round to other decimal places as you mentioned, simply multiply the initial value by 10, or 100, etc. use the same sort of code, and then divide the result by the same number and you'll get the same result at whatever decimal place you want.
Here's an example for rounding at an arbitrary precision.
double valueToRound = GetTheValueFromSomewhere();
int decimalPrecisionAtWhichToRound = 0;
double scale = 10^decimalPrecisionAtWhichToRound;
double tmp = valueToRound * scale;
tmp = (double)((int)(tmp + 0.5));
double roundedValue = tmp / scale;
So, if decimalPrecisionAtWhichToRound is set to 0 as in the above it'll round to the nearest whole integer. 1.4 will round to 1.0. 1.5 will round to 2.0.
If you set decimalPrecisionAtWhichToRound to 1, it would round to the nearest tenth. 1.45 would round to 1.5 and 1.43 would round to 1.4.
You need to first understand how to do rounding on paper, without someone showing you the code to do it. Write down some numbers and figure out how to round them.
To round to a specific decimal position you add half the value of that position and then truncate. Ie, 1.67 + 0.05 = 1.72 then truncate to 1.7.
But there are two tricky things in programming that aren't there when you do it on paper:
Knowing how to truncate -- There are several ways to do it while programming, but they're non-trivial.
Dealing with the fact that floating-point numbers are imprecise. Ie, there is no exact representation of, say, 1.7, but rather the two closest numbers are apt to be something like 1.69998 and 1.700001
For truncating the trick of multiplying the number by the appropriate power of 10 to produce an integer works pretty well. Eg, (1.67 + 0.05) * 10 = 17.2, then convert to int to get 17, then convert back to float and divide by 10 to get 1.7 (more or less). Or (if you're printing or displaying the value) just format the integer number with the decimal point inserted. (By formatting the integer value you don't have to deal with the problem of imprecise floating point representations.)
If you want to suppress trailing zeros it gets a bit trickier and you probably have to actually write some code -- format the number, then scan backwards and take off any trailing zeros up to the decimal point. (And take the decimal point too, if you wish.)
float number=17.125;
NSNumberFormatter *format = [[NSNumberFormatter alloc]init];
[format setNumberStyle:NSNumberFormatterDecimalStyle];
[format setRoundingMode:NSNumberFormatterRoundHalfUp];
[format setMaximumFractionDigits:2];
NSString *temp = [format stringFromNumber:[NSNumber numberWithFloat:number]];
NSLog(#"%#",temp);
double myNumber = 7.99;
NSString *formattedNumber = [NSString stringWithFormat:#"%.*f",
fmod(round(myNumber * 10), 10) ? 1 : 0, myNumber];
I have found multiple ways to set a sliders value based on ints but I need to increment according to a float .1.
float value = [_xSlider value] *.1;
[_xSlider setValue: value/.1];
But I get a large float number not just 0.3.
I found the answer.
I needed to put the number of decimal places I wanted in front of the decimal then act like it was an int, then push the numbers back onto the other side of the decimal.
int value = [_xSlider value] * 10;
[_xSlider setValue: value * .1];