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];
Related
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.
I am having an issue with using a float in a UILabel.
float doubleNum;
floatNum = 10 / 20;
cashLabel.text = [[NSString alloc] initWithFormat:#"%f", floatNum];
If I use "floatNum = 10 / 10;" it correctly returns "1.000000000", however, if I put in "floatNum = 10 / 20" it returns "0.0000000". I have tried about everything I know and it does not work. I know it's a dumb mistake, but I can't figure it out.
Happy Holidays. :)
You need to cast one of the integer's to a float.
Try replacing the divisional line with:
float floatNum = (float) 10 / 20;
and you should get the correct answer.
Or if possible just use floats in your division:
float floatNum = 10.0f / 20.0f;
should also work
The issue here is that you are assigning floatNum the result of dividing one INTEGER by another. The result of 10 / 20 is indeed 0 and as a float, it appears as 0.0000000. In order to obtain a proper result, you need to either use a cast type to turn it into a float or add a .0 to one of the numbers. In division, if one of the numbers is a float (which is easily done by just adding a .0 to one of them), the result will be a float as well.
Normally, C performs "integer division" (basically, division without the remainder -- 10/3 is 3R1, so it yields 3).
When you type floatNum = 10/20, it does 10/20 = 0 (remainder 10).
To fix this, you have to tell the program that you're giving it floating point numbers. Try using:
floatNum = 10.0 / 20,
floatNum = 10 / 20.0, or
floatNum = float (10 / 20).
All of those should work.
try floatNum = 10.0f/20.0f (i.e. make sure the calculation is being done with floats rather than ints)
Consider following example to understand how floats work:
float a = 1/120;
float b = 1.0/120;
float c = 1.0/120.0;
float d = 1.0f/120.0f;
NSLog(#"Value of A:%f B:%f C:%f D:%f",a,b,c,d);
Output: Value of A:0.000000 B:0.008333 C:0.008333 D:0.008333
For float variable a : int / int yields integer which you are assigning to float and printing it so 0.0000000
For float variable b: float / int yields float, assigning to float and printing it 0.008333
For float variable c: float / float yields float, so 0.008333
Last one is more precise float. Previous ones are of type double: all floating point values are stored as double data types unless the value is followed by an 'f' to specifically specify a float rather than as a double.
Change your code to:
float floatNum;
floatNum = 10.0f / 20.0f;
cashLabel.text = [[NSString alloc] initWithFormat:#"%f", floatNum];
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];
How do I round a float to three decimal places?
I have this:
(round(1000.0f * currentHue) / 1000.0f)
Which rounds it to 3 decimal places but leaves a bunch of 0s on the end.
Most numbers can't be represented precisely with floating point. If you need precision that bad pick appropriate scientific library. If you want to print it nicely, use formatters -
NSLog(#"%.3f", currentHue);
This is a fairly hairy way to do it... But it should work.
float x = [[NSString stringWithFormat:#"%1.3f",(round(1000.0f * currentHue) / 1000.0f)] floatValue];
How might I round a float to the nearest integer in Objective-C:
Example:
float f = 45.698f;
int rounded = _______;
NSLog(#"the rounded float is %i",rounded);
should print "the rounded float is 46"
Use the C standard function family round(). roundf() for float, round() for double, and roundl() for long double. You can then cast the result to the integer type of your choice.
The recommended way is in this answer: https://stackoverflow.com/a/4702539/308315
Original answer:
cast it to an int after adding 0.5.
So
NSLog (#"the rounded float is %i", (int) (f + 0.5));
Edit: the way you asked for:
int rounded = (f + 0.5);
NSLog (#"the rounded float is %i", rounded);
For round float to nearest integer use roundf()
roundf(3.2) // 3
roundf(3.6) // 4
You can also use ceil() function for always get upper value from float.
ceil(3.2) // 4
ceil(3.6) // 4
And for lowest value floor()
floorf(3.2) //3
floorf(3.6) //3
The easiest way to round a float in objective-c is lroundf:
float yourFloat = 3.14;
int roundedFloat = lroundf(yourFloat);
NSLog(#"%d",roundedFloat);
Check the manual page for rint()
If in case you want round float value in integer below is the simple method for rounding the float value in objective C.
int roundedValue = roundf(Your float value);
let's do tried and checkout
//Your Number to Round (can be predefined or whatever you need it to be)
float numberToRound = 1.12345;
float min = ([ [[NSString alloc]initWithFormat:#"%.0f",numberToRound] floatValue]);
float max = min + 1;
float maxdif = max - numberToRound;
if (maxdif > .5) {
numberToRound = min;
}else{
numberToRound = max;
}
//numberToRound will now equal it's closest whole number (in this case, it's 1)