In this case:
float a = 0.99999f;
int b = 1000;
int c = a + b;
In result c = 1001. I discovered that it happens because b is converted to float (specific for iOS), then a + b doesn't have enough precision for 1000.9999 and (why?) is rounded to higher value. If a is 0.999f we get c = 1000 - theoretically correct behavior.
So my question is why float number is rounded to higher value? Where this behavior (or convention) is described?
I tested this on iPhone Simulator, Apple LLVM 4.2 compiler.
In int c = a + b, the integer b is converted to a float first, then 2 floating point
numbers are added, and the result is truncated to an integer.
The default floating point rounding mode is FE_TONEAREST, which means that the result
of the addition
0.99999f + 1000f
is the nearest number that can be represented as a float, and that is the number 1001f. This float is then truncated to the integer c = 1001.
If you change the rounding mode
#include <fenv.h>
fesetround(FE_DOWNWARD);
then the result of the addition is rounded downward (approximately 1000.99993f) and you would get c = 1000.
The reason is that when you add 1000 you get 8 total decimal digits of precision, but IEEE float is only supports 7 digits.
Related
hi I have a big doubt of using floats for divisions,
here the test code::
- (void)chuza
{
float option1 = 0;
float option2 = 0;
option1 = 100/50;
option2 = 50/100;
NSLog(#"option1 :: %f",option1);
NSLog(#"option2 :: %f",option2);
}
So,
I get:
2012-10-19 16:21:45.405 Project[2023:14f03] option1 :: 2.000000
2012-10-19 16:21:45.405 Project[2023:14f03] option2 :: 0.000000
so my question is:
why would the float like when
denominator > numerator
but doesnt like
numerator > denominator ??
Thanks!
That looks a lot like integer division to me since you are dividing two integer numbers. When dividing two integers (in most if not all programming languages) ignores the remainder of the division.
Think of how that line is being evaluated. First the integer division is calculated and then the integer result is cast to a float to be stored in the float variable.
Note that integers don't store any decimals at all so even 0.999 as an integer would be 0. It's not a problem about rounding.
It's also not about the denominator being bigger than the numerator. Try dividing 100/30 and the result will be 3, not 3.33333.. as it would be for float division.
You can solve this by casting the numbers to floats or making sure they are float numbers.
Casting
option2 = ((float)50/(float)100);
Dividing floats
option2 = 50.0f/100.0f;
Neither 100 nor 50 are floats, they're integers. That means you're doing integer division and assigning the result to a floating point variable. Since integer division truncates (throws away fractional results), 50/100 yields 0. 100 is divisible by 50, which which is why you get the 2.000000 result. Try your program with 75 and 100, and you'll see there's more at work than just which direction the division goes.
It's treating your numbers as integers, performing the division and then putting the result into a float.
Try this option2 = 50.0f/100;
Im obtaining an int value from UITextField [self.dbRef.text intValue];
I want to then format that value so I can add a decimal place that precceds the number ie. If [self.dbRef.text intValue]; returns 4 i need that value to be 0.04
So far I have tried various ways including
float Y = ([self.dbRef.text intValue]/100);
slice.value = Y;
NSLog(#"float Y value = %f",Y);
returns zero
NSString* formatedTotalApplianceString = [NSString stringWithFormat:#"0.%#", self.dbRef.text];
NSLog(#"formated string = %#",formatedTotalApplianceString);
int totalAppliances = [formatedTotalApplianceString intValue];
NSLog(#"Resulting int value = %d",[formatedTotalApplianceString intValue]);
slice.value = totalAppliances;
NSLog(#"total appliances int value = %d",totalAppliances);
returns zero
You're doing an integer division, so the 0 value is correct in that context as integers cannot represent fractions (unless you're doing fixed point arithmetics, but that's a different can of worms). You need to do a floating point division, for example:
float Y = ([self.dbRef.text floatValue]/100.0f);
Either the [self.dbRef.text floatValue] or the 100.0f will turn this into a float division, because if the other side would be an int it would automatically get casted to a float. But the "best" way is to have both values of the same type.
Change
float Y = [self.dbRef.text intValue]/100;
to
float Y = ((float)[self.dbRef.text intValue])/100;
in your first variant.
Dividing int by int returns you int result even if then you assign it to float. 4/100 = 0 in such case.
The problem with [self.dbRef.text intValue]/100 is that it's an integer division. It drops the fraction. One way to work around it is to divide by 100.0:
[self.dbRef.text intValue]/100.0
However, this is not the most efficient way of doing it if all you need is adding a zero in front of a fraction: you could avoid float altogether by padding your printed int to two positions with leading zeros:
// If text is 4, the code below prints 0.04
NSLog(#"0.%02d", [self.dbRef.text intValue]);
The first code returns zero because you are performing an integer division, which produces an integer result. You should cast the value to a float.
The second code also returns zero because you're asking for the intValue of a floating point value. So the decimal part will be discarded.
NSString has also a floatValue method, you should use it to get a floating value. Once divided by 100 you will still have a floating point value (in a division if the quotient or the dividend is a float and the other an integer, the integer gets promoted to float):
float Y = ([self.dbRef.text floatValue]/100);
slice.value = Y;
I am developing a program that solves a system of equations. When it gives me the results, it is like: "x1= 1,36842". I'd like to get the fraction of that "1,36842", so I wrote this code.
procedure TForm1.Button1Click(Sender: TObject);
var numero,s:string;
a,intpart,fracpart,frazfatta:double;
y,i,mcd,x,nume,denomin,R:integer;
begin
a:=StrToFloat(Edit1.Text); //get the value of a
IntPart := Trunc(a); // here I get the numerator and the denominator
FracPart := a-Trunc(a);
Edit2.Text:=FloatToStr(FracPart);
numero:='1';
for i:= 1 to (length(Edit2.Text)-2) do
begin
numero:=numero+'0';
end; //in this loop it creates a string that has many 0 as the length of the denominator
Edit3.text:=FloatToStr(IntPart);
y:=StrToInt(numero);
x:=StrToInt(Edit3.Text);
while y <> 0 do
begin
R:= x mod y;
x:=y;
y:=R;
end;
mcd:=x; //at the end of this loop I have the greatest common divisor
nume:= StrToInt(Edit3.Text) div mcd;
denomin:= StrToInt(numero) div mcd;
Memo1.Lines.Add('fraction: '+IntToStr(nume)+'/'+IntToStr(denomin));
end;
It doesn't work correctly because the fraction that it gives to me is wrong. Could anyone help me please?
Your code cannot work because you are using binary floating point. And binary floating point types cannot represent the decimal numbers that you are trying to represent. Representable binary floating point numbers are of the form s2e where s is the significand and e is the exponent. So, for example, you cannot represent 0.1 as a binary floating point value.
The most obvious solution is to perform the calculation using integer arithmetic. Don't call StrToFloat at all. Don't touch floating point arithmetic. Parse the input string yourself. Locate the decimal point. Use the number of digits that follow to work out the decimal scale. Strip off any leading or trailing zeros. And do the rest using integer arithmetic.
As an example, suppose the input is '2.79'. Convert that, by processing the text, into numerator and denominator variables
Numerator := 279;
Denominator := 100;
Obviously you'd have to code string parsing routines rather than use integer literals, but that is routine.
Finally, complete the problem by finding the gcd of these two integers.
The bottom line is that to represent and operate on decimal data you need a decimal algorithm. And that excludes binary floating point.
I recommend defining a function GreaterCommonDivisor function first (wiki reference)
This is going to be Java/C like code since I'm not familiar with Delphi
let
float x = inputnum // where inputnum is a float
// eg. x = 123.56
Then, multiplying
int n = 1;
while(decimalpart != 0){// or cast int and check if equal-> (int)x == x
x = x * 10;
decimalpart = x % 1;
// or a function getting the decimal part if the cast does work
n *= 10;
}
// running eg. x = 123.56 now x = 12356
// n = 100
Then you should have (float)x/n == inputnum at this point eg. (12356/100 == 123.56)
This mean you have a fraction that may not be simpified at this point. All you do now is implement and use the GCD function
int gcd = GreaterCommonDivisor(x, n);
// GreaterCommonDivisor(12356, 100) returns 4
// therefore for correct implementation gcd = 4
x /= gcd; // 12356 / 4 = 3089
n /= gcd; // 100 / 4 = 25
This should be quick and simple to implement, but:
Major Pitfalls:
Float must be terminating. For example expected value for 0.333333333333333333 won't be rounded to 1/3
Float * n <= max_int_value, otherwise there will be a overflow, there are work around this, but there may be another solutions more fitting to these larger numbers
Continued fractions can be used to find good rational approximations to real numbers. Here's an implementation in JavaScript, I'm sure it's trivial to port to Delphi:
function float2rat(x) {
var tolerance = 1.0E-6;
var h1=1; var h2=0;
var k1=0; var k2=1;
var b = x;
do {
var a = Math.floor(b);
var aux = h1; h1 = a*h1+h2; h2 = aux;
aux = k1; k1 = a*k1+k2; k2 = aux;
b = 1/(b-a);
} while (Math.abs(x-h1/k1) > x*tolerance);
return h1+"/"+k1;
}
For example, 1.36842 is converted into 26/19.
You can find a live demo and more information about this algorithm on my blog.
#Joni
I tried 1/2 and the result was a "division by zero" error;
I correct the loop adding:
if b - a = 0 then BREAK;
To avoid
b:= 1 / (b - a);
I have a number in actionscript, arrived at via some arbitrary math:
var value:Number = 45 * (1 - (1 /3));
trace(value);//30.00000000004
Now, I would like to take the ceiling of this number, except in cases where the amount it is greater than the next lower integer is smaller than some epsilon. In the above example, I really want to round to 30, but only in the case where I know I'm getting a rounding error:
Math.ceil(value); //I want 30, but get 31
Math.ceil(30.1); //In this case, it's reasonable to get 31
Is there an elegant way to truncate a Number in actionscript? Or easily discard any part of the number that is less than some epsilon?
Is this method is of any help to you?
var precision:int = 4;
var isActualCeilingValRequred:Boolean;
var thresholdValForCeiling:int = 100;
private function getCeilingValue(num:Number):Number
{
var tempNum = num * Math.pow(10, precision);
var decimalVal = tempNum % Math.pow(10, precision);
if(decimalVal < thresholdValForCeiling) {
return Math.floor(num);
} else {
return Math.Ceil(num);
}
}
var value:Number = 45 * (1 - (1 /3));
trace(value);//30.00000000004
// Play with arbitraryPrecision until you are satisfied with
// the accuracy of your results
var arbitraryPrecision:int = 3;
var fixed:Number = value.toFixed(arbitraryPrecision);
trace(Math.ceil(fixed));
The basic way to round a number to a specified number of fractional digits is to multiply the number to 10^DIGITS to shift the decimal point DIGITS digits to the left, perform the rounding, and divide by the same 10^DIGITS to shift the decimal point back to the right.
var value:Number = 45 * (1 - (1 / 3));
trace(value); // 30.000000000000004
trace(Math.ceil(value)); // 31
// Round the number to 13 decimal digits.
const POWER:Number = 1e13;
value = Math.round(value * POWER) / POWER;
trace(value); // 30
// Compute number's ceiling.
value = Math.ceil(value);
trace(value); // 30`
It works for your example, but there's a big gotcha. If you change your value to be 450 * (1 - (1 / 3));, your original problem will appear again. Now to get rid of it, you would have to round to 12 decimal digits. Basically, the significand of a double-precision format (Number) can hold about 15 significant digits. This means as the value increases by a factor of ten, the decimal points moves to the left and that last "4" digit you want to get rid of becomes closer and closer to the decimal point. So the code becomes more complicated.
var value:Number = 450 * (1 - (1 / 3));
trace(value); // 30.000000000000004
trace(Math.ceil(value)); // 31
var exp:Number = Math.floor(Math.log(Math.abs(value)) * Math.LOG10E);
trace('exp=' + exp); // exp=2
const POWER:Number = Math.pow(10, 14 - exp);
value *= POWER;
trace(value); // 300000000000000.06
value = Math.round(value);
trace(value); // 300000000000000
value /= POWER;
trace(value); // 300
As you can see, it now works regardless of the value's magnitude.
First, I find the number's exponent by taking a base-10 logarithm of the number's absolute value, then rounding it down. If you calculate a = value * Math.pow(10, exp);, then value could be represented as a * 10^b, where (1 ≤ |a| < 10), known as normalized scientific notation. But that's not what we're doing here. Now that we know how many digits are on the left of the decimal point, we will shift the decimal point right, but not too far, to keep one 0 and this error digit we want to get rid of, on the right side of the decimal point. So, multiply by 10^(14-exp), round, then divide by the same power.
I'm trying to generate a random number that's between 0 and 1. I keep reading about arc4random(), but there isn't any information about getting a float from it. How do I do this?
Random value in [0, 1[ (including 0, excluding 1):
double val = ((double)arc4random() / UINT32_MAX);
A bit more details here.
Actual range is [0, 0.999999999767169356], as upper bound is (double)0xFFFFFFFF / 0x100000000.
// Seed (only once)
srand48(time(0));
double x = drand48();
// Swift version
// Seed (only once)
srand48(Int(Date().timeIntervalSince1970))
let x = drand48()
The drand48() and erand48() functions return non-negative, double-precision, floating-point values, uniformly distributed over the interval [0.0 , 1.0].
For Swift 4.2+ see: https://stackoverflow.com/a/50733095/1033581
Below are recommendations for correct uniformity and optimal precision for ObjC and Swift 4.1.
32 bits precision (Optimal for Float)
Uniform random value in [0, 1] (including 0.0 and 1.0), up to 32 bits precision:
Obj-C:
float val = (float)arc4random() / UINT32_MAX;
Swift:
let val = Float(arc4random()) / Float(UInt32.max)
It's optimal for:
a Float (or Float32) which has a significand precision of 24 bits for its mantissa
48 bits precision (discouraged)
It's easy to achieve 48 bits precision with drand48 (which uses arc4random_buf under the hood). But note that drand48 has flaws because of the seed requirement and also for being suboptimal to randomize all 52 bits of Double mantissa.
Uniform random value in [0, 1], 48 bits precision:
Swift:
// seed (only needed once)
srand48(Int(Date.timeIntervalSinceReferenceDate))
// random Double value
let val = drand48()
64 bits precision (Optimal for Double and Float80)
Uniform random value in [0, 1] (including 0.0 and 1.0), up to 64 bits precision:
Swift, using two calls to arc4random:
let arc4random64 = UInt64(arc4random()) << 32 &+ UInt64(arc4random())
let val = Float80(arc4random64) / Float80(UInt64.max)
Swift, using one call to arc4random_buf:
var arc4random64: UInt64 = 0
arc4random_buf(&arc4random64, MemoryLayout.size(ofValue: arc4random64))
let val = Float80(arc4random64) / Float80(UInt64.max)
It's optimal for:
a Double (or Float64) which has a significand precision of 52 bits for its mantissa
a Float80 which has a significand precision of 64 bits for its mantissa
Notes
Comparisons with other methods
Answers where the range is excluding one of the bounds (0 or 1) likely suffer from a uniformity bias and should be avoided.
using arc4random(), best precision is 1 / 0xFFFFFFFF (UINT32_MAX)
using arc4random_uniform(), best precision is 1 / 0xFFFFFFFE (UINT32_MAX-1)
using rand() (secretly using arc4random), best precision is 1 / 0x7FFFFFFF (RAND_MAX)
using random() (secretly using arc4random), best precision is 1 / 0x7FFFFFFF (RAND_MAX)
It's mathematically impossible to achieve better than 32 bits precision with a single call to arc4random, arc4random_uniform, rand or random. So our above 32 bits and 64 bits solutions should be the best we can achieve.
This function works for negative float ranges as well:
float randomFloat(float Min, float Max){
return ((arc4random()%RAND_MAX)/(RAND_MAX*1.0))*(Max-Min)+Min;
}
Swift 4.2+
Swift 4.2 adds native support for a random value in a Range:
let x = Float.random(in: 0.0...1.0)
let y = Double.random(in: 0.0...1.0)
let z = Float80.random(in: 0.0...1.0)
Doc:
random(in range: ClosedRange<Float>)
random(in range: Range<Float>)
random(in range: ClosedRange<Double>)
random(in range: Range<Double>)
random(in range: ClosedRange<Float80>)
random(in range: Range<Float80>)
(float)rand() / RAND_MAX
The previous post stating "rand()" alone was incorrect.
This is the correct way to use rand().
This will create a number between 0 -> 1
BSD docs:
The rand() function computes a sequence of pseudo-random integers in the
range of 0 to RAND_MAX (as defined by the header file "stdlib.h").
This is extension for Float random number Swift 3.1
// MARK: Float Extension
public extension Float {
/// Returns a random floating point number between 0.0 and 1.0, inclusive.
public static var random: Float {
return Float(arc4random()) / Float(UInt32.max))
}
/// Random float between 0 and n-1.
///
/// - Parameter n: Interval max
/// - Returns: Returns a random float point number between 0 and n max
public static func random(min: Float, max: Float) -> Float {
return Float.random * (max - min) + min
}
}
Swift 4.2
Swift 4.2 has included a native and fairly full-featured random number API in the standard library. (Swift Evolution proposal SE-0202)
let intBetween0to9 = Int.random(in: 0...9)
let doubleBetween0to1 = Double.random(in: 0...1)
All number types have the static random(in:) function which takes the range and returns the random number in the given range.
Use this to avoid problems with upper bound of arc4random()
u_int32_t upper_bound = 1000000;
float r = arc4random_uniform(upper_bound)*1.0/upper_bound;
Note that it is applicable for MAC_10_7, IPHONE_4_3 and higher.
arc4random has a range up to 0x100000000 (4294967296)
This is another good option to generate random numbers between 0 to 1:
srand48(time(0)); // pseudo-random number initializer.
double r = drand48();
float x = arc4random() % 11 * 0.1;
That produces a random float bewteen 0 and 1.
More info here
rand()
by default produces a random number(float) between 0 and 1.