A float problem in iOS when indexPath.row=1
Float32 value=0.05*indexPath.row;
it returns value=0.0500000007 but it should be 0.05. How to fix this??
Welcome to floating point errors.
The easiest way to fix this is just to round the number after multiplying.
You can read more about why this is happening on this random website I found on Google.
Related
Is this the 1 billionth ugly/hamming number?
62565096724471903888424537973014890491686968126921250076541212862080934425144389
76692222667734743108165348546009548371249535465997230641841310549077830079108427
08520497989078343041081429889246063472775181069303596625038985214292236784430583
66046734494015674435358781857279355148950650629382822451696203426871312216858487
7816068576714140173718
Does anyone have code to share that can verify this? Thanks!
This SO answer shows a code capable of calculating it.
The test entry on ideone.com takes 1.1 0.05 sec for 109 (2016-08-18: main speedup due to usage of Int instead of the default Integer where possible, even on 32-bit; additional 20% thanks to the tweak suggested by #GordonBGood, bringing band size complexity down to O(n1/3)).
It gives the answer as ((1334,335,404),"6.21607575556559E+843"), i.e.
21334 * 3335 * 5404 ≈ 6.21607575556559 * 10843.
(coincidentally, only two last digits in the fractional number above are incorrect).
This also means, of course, that there are 404 zeroes at the end of this number, and that it has 844 digits in total. So no, the number you show isn't it.
Exact answer:
6216075755565244861630816332872072003947056519089652706591632409642337022002753141824417540777256732780370172616615291935540418620025524916729500086831454711313694078635504004160312872951788703647948382456091072701600790562071797590306654765882256990391763887850141154482249915927439184562828227449023750262318234797192076792208033475638322151983772515798004125909334741121595323950448656375104457026997424772966917441779406172736975588556800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
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.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Trouble with floats in Objective-C
I have broken this problem down to about as simple as i can get it. Feel free to try the same thing and tell me if you get the same error and what solution you might have. I have already tried it on several computers.
float total = 200000.0f + 154196.8f;
NSLog(#"total: %f", total);
The output is:
total: 354196.812500
If anyone has any sort of logical explanation, feel free to share it.
I'd suggest you brush up on your floats
http://www.altdevblogaday.com/2012/05/20/thats-not-normalthe-performance-of-odd-floats/
If you need higher precision use a double.
Additionally http://randomascii.wordpress.com/2012/03/08/float-precisionfrom-zero-to-100-digits-2/
See What Every Programmer Should Know About Floating-Point Arithmetic for all the deep understanding. The short answer is that all floating point representations have limitations on their precision, and that things that can be expressed in a small number of digits in decimal may not be expressible in a small number of digits in binary (and specifically not in floating point formats).
Note that while double can improve things, it is no panacea. It is quite common to have small rounding errors, even with double. You may easily get 1.99999999 when you expect 2.
Hint:
long double total = 200000.0 + 154196.8;
NSLog(#"total: %Lf", total);
On my machine prints the correct value.
A 32 bit floating point has a 23 bit mantissa, the closest value is 0.5+0.25+0.125.
You should use more bits to get the correct representation.
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.
I've a small problem and I can't find a solution!
My code is (this is only a sample code, but my original code do something like this):
float x = [#"2.45" floatValue];
for(int i=0; i<100; i++)
x += 0.22;
NSLog(#"%f", x);
the output is 52.450001 and not 52.450000 !
I don't know because this happens!
Thanks for any help!
~SOLVED~
Thanks to everybody! Yes, I've solved with the double type!
Floats are a number representation with a certain precision. Not every value can be represented in this format. See here as well.
You can easily think of why this would be the case: there is an unlimited number of number just in the intervall (1..1), but a float only has a limited number of bits to represent all numbers in (-MAXFLOAT..MAXFLOAT).
More aptly put: in a 32bit integer representation there is a countable number of integers to be represented, But there is an infinite innumerable number of real values that cannot be fully represented in a limited representation of 32 or 64bit. Therefore there not only is a limit to the highest and lowest representable real value, but also to the accuracy.
So why is a number that has little digits after the floating point affected? Because the representation is based on a binary system instead of a decimal, making other numbers easily represented then the decimal ones.
See http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems
Floating point numbers can not always be represented easily by computers. This leads to inaccuracy in some digits.
It's like me asking you what 1/3 is in decimal. No matter how hard you try, you're not going to be able to tell me what it is because decimal can't accurately describe that number.
Floats can't accurately describe some decimal numbers.