I'm struggling with making sense of how to return the changeDue for my assignment. Trying to revise my incorrect code for class in prep for intro to programming final.
all I am trying to do is: create a method called quarters(). When I pass any double value (ChangeDue) into the method, I want to know precisely how many quarters there are as well as the partial quarter change returned.
Original code:
func getChange(Quarters: Double) -> Double {
var Change = Quarters
return Change;
}
var Quarters = 0.72;
var ChangeDue = getChange(Quarters / .25);
print(ChangeDue)
Slightly revised code which I seem to have made worse:
class changeDue {
var = quarters(.72)
func changeDue(Quarters: Double) {
var Change = Quarters
changeDue = changeDue - (quarters*.25)
}
var ChangeDue = getChange(int / .25);
print(changeDue)
}
notes/Feedback:
create a method called quarters(). When I pass any double value (ChangeDue) into the method, I want to know precisely how many quarters there are as well as the partial quarter change returned.
Create a class level variable, changeDue. This is where you will set your test input e.g. .78, 2.15.
In your method, calculate the number of quarters as the integer of changeDue/.25
Print the number of quarters.
Now you need the revised change after quarters are removed. changeDue=changeDue - (quarters*.25)
quarters = the integer of changedue/.25
changeDue is now = to the previous changeDue - (quarters times .25)
quarters(.72)
the integer of .72/.25 = 2
changedue=.72-(2 x .25) or .72 - .50 =.12
Print changeDue.
Any help would be appreciated. I've been working on this for longer than I want to admit.
Hint 1: Do not work with Double or fractional amounts. Turn dollars into pennies by multiplying everything by 100 before you start. Now you can do everything with integer arithmetic. After you get the answer, you can always divide by 100 to turn it back into dollars, if desired.
Hint 2: Do you know about the % operator? It tells you the remainder after a division.
I don't want to write your code for you (it's you who are being tested, not me, after all), but I'll just demonstrate with a different example:
51 / 7 is 7, because integer division throws away the remainder.
Sure, 7x7 is 49, with something left over. But what?
Answer: 51 % 7 is 2. Do you see?
Related
I am trying to get remainder using swift's truncatingRemainder(dividingBy:) method.
But I am getting a non zero remainder even if value I am using is completely divisible by deviser. I have tried number of solutions available here but none worked.
P.S. values I am using are Double (Tried Float also).
Here is my code.
let price = 0.5
let tick = 0.05
let remainder = price.truncatingRemainder(dividingBy: tick)
if remainder != 0 {
return "Price should be in multiple of tick"
}
I am getting 0.049999999999999975 as remainder which is clearly not the expected result.
As usual (see https://floating-point-gui.de), this is caused by the way numbers are stored in a computer.
According to the docs, this is what we expect
let price = //
let tick = //
let r = price.truncatingRemainder(dividingBy: tick)
let q = (price/tick).rounded(.towardZero)
tick*q+r == price // should be true
In the case where it looks to your eye as if tick evenly divides price, everything depends on the inner storage system. For example, if price is 0.4 and tick is 0.04, then r is vanishingly close to zero (as you expect) and the last statement is true.
But when price is 0.5 and tick is 0.05, there is a tiny discrepancy due to the way the numbers are stored, and we end up with this odd situation where r, instead of being vanishingly close to zero, is vanishing close to tick! And of course the last statement is then false.
You'll just have to compensate in your code. Clearly the remainder cannot be the divisor, so if the remainder is vanishingly close to the divisor (within some epsilon), you'll just have to disregard it and call it zero.
You could file a bug on this but I doubt that much can be done about it.
Okay, I put in a query about this and got back that it behaves as intended, as I suspected. The reply (from Stephen Canon) was:
That's the correct behavior. 0.05 is a Double with the value 0.05000000000000000277555756156289135105907917022705078125. Dividing 0.5 by that value in exact arithmetic gives 9 with a remainder of 0.04999999999999997501998194593397784046828746795654296875, which is exactly the result you're seeing.
The only rounding error that occurs in your example is in the division price/tick, which rounds up to 10 before your .rounded(.towardZero) has a chance to take effect. We'll add an API to let you do something like price.divided(by: tick, rounding: .towardZero) at some point, which will eliminate this rounding, but the behavior of truncatingRemainder is precisely as intended.
You really want to have either a decimal type (also on the list of things to do) or to scale the problem by a power of ten so that your divisor become exact:
1> let price = 50.0
price: Double = 50
2> let tick = 5.0
tick: Double = 5
3> let r = price.truncatingRemainder(dividingBy: tick)
r: Double = 0
Basically I would like to know if it is possible to get the exponent value in a number, ex:
number = 2.6e3
I want to get the value 3 of the exponent. I have been searching for quite a while now and have not found the answer to this. I am new to programming so I may not know exactly what to look for (which methods, etc).
Any help is much appreciated! Thanks!
Assuming I am interpreting your question correctly this is what you want to do:
B = A^X where A and B are known values. Solve for X.
1000 = 10^X (In this case, X = 3.)
The below code will work for any base. It requires either Foundation or UIKit. The function arguments "value" and "base" are B, A respectively. Try the code out in the Xcode Playground!
func getExponentForValueAndBase(value: Double, base: Double) -> Double {
return log(value)/log(base)
}
getExponentForValueAndBase(1000, base: 10) // = 3
Assuming this is your question: Given a number as an integer, find the interger value of the log base 10 of it.
import Foundation
func log(Int number) -> Int
{
return floor(log10(number))
}
Using Swift, I am trying to figure out how to use arc4random_uniform to return a number like 37.7. The guidance I must abide by is I must do it in a function, the random double must be between 0 - 300. I have been able to build a function that randomly returns doubles between the range but can't find anything that will lead me to outputting random non whole numbers
//function to randomly generate a double number like 105.3
func makeRandDbl() -> Double {
let randGenerator: Double = Double(arc4random_uniform(301))
print(randGenerator)
return randGenerator
}
makeRandDb()
To generate a Double in the range 0.0 to 300.0 (with one digit following the decimal):
Double(arc4random_uniform(3001))/10.0
You can extend this to more decimal places. For two decimal places (0.00 to 300.00):
Double(arc4random_uniform(30001))/100.0
For three decimal places (0.000 to 300.000):
Double(arc4random_uniform(300001))/1000.0
This has the advantage of being able to actually generate whole values. In the first case 10% of the numbers will be whole. In the second case 1% of the numbers will be whole. And in the third, 0.1% of the numbers will be whole.
This is your function, I believe:
extension Double {
/// Generates a random `Double` within `0.0...1.0`
public static func random() -> Double {
return random(0.0...1.0)
}
/// Generates a random `Double` inside of the closed interval.
public static func random(interval: ClosedInterval<Double>) -> Double {
return interval.start + (interval.end - interval.start) * (Double(arc4random()) / Double(UInt32.max))
}
}
Usage example:
Double.random(0...300)
It is taken from RandomKit library - it looks very useful for various purposes.
One approach would be to convert the result of arc4random_uniform to double, divide the result by UInt32.max, and then multiply the result by 300.
let rand = 300 * Double(arc4random_uniform(UInt32.max)) / Double(UInt32.max)
This would produce a value between 0 and 300, inclusive. The number of possible values that you are going to get is UInt32.max.
I’m a beginner in programming and playing around with the arc4random_uniform() function in Swift. The program I’m making so far generates a random number from 1-10 regenerated by a UIButton. However, I want the variable ’highest' that gets initialised to the random number to update if the next generated number is larger than the one currently held in it. For example the random number is 6 which is stored in highest and if the next number is 8 highest becomes 8. I don't know how to go about this. I have connected the UIButton to an IBAction function and have the following code:
var randomValue = arc4random_uniform(11) + 1
highest = Int(randomValue)
if (Int(randomValue) < highest) {
// Don’t know what to do
}
Initialise highest to 0
Every time you generate a new random number, replace the value of highest with the higher of the two numbers
highest = max(highest, randomValue)
The max() function is part of the Swift standard library and returns the larger of the two passed in vales.
edited to add
Here's a playground showing this with a bit more detail, including casting of types:
var highest: Int = 0
func random() -> Int {
let r = arc4random_uniform(10) + 1
return Int(r)
}
var randomValue = random()
highest = max(highest, randomValue)
You can see that multiple calls persist the highest value.
I have a very simple function to convert temperature from ˚C TO ˚K.
func convertKelvinToCelsius(temp:Double) ->Double {
return temp - 273.15
}
And I have a unit test to drive this function. This is where the problem is:
func testKelvinToCelsius(){
var check1 = conv.convertKelvinToCelsius(200.00) // -73.149999999999977
var check2 = 200.00 - 273.15 // -73.149999999999977
var check3 = Double(-73.15) // -73.150000000000006
//Passes
XCTAssert(conv.convertKelvinToCelsius(200.00).description == Double(-73.15).description, "Shoud convert from celsius kelvin")
//Fails
XCTAssert(conv.convertKelvinToCelsius(200.00) == Double(-73.15), "Shoud convert from celsius kelvin")
}
When you add a breakpoint and check the values of check1, check2 and check3, they are very interesting:
check1 Double -73.149999999999977
check2 Double -73.149999999999977
check3 Double -73.150000000000006
Questions:
Why does Swift return different values for check1/check2 and check3
How can I get the second test to pass, because writing it like I did the test1 smells. Why should I have to convert Doubles to Strings to be able to compare them?
Finally, when I println check1, check2 and check3, they all print to be '-73.15'. Why? Why not print accurately, and not confuse the programmers!?
To Reproduce:
Just type 200 - 273.15 == -73.15 in you playground and watch it go false!!
This is expected behavior for floating point values. They cannot be 100% accurately represented.
You can use the XCTAssertEqualWithAccuracy function to assert floating point values are within a given range of each other.
The reason println prints the same value for all is because it internally rounds them to two decimals (I assume).
This is not a Swift specific issue, this is related to the fact how decimal numbers are created in computers and what is their precision. You will need to work with DBL_EPSILON.
Swift, like most languages, uses binary floating point numbers.
With binary floating point numbers, some numbers can be represented exactly, but most can't. What can be represented exactly are integers unless they are very large (for example, 100000000000000.0 is fine), and such integers multiplied or divided by powers of two (7.375 is fine, it is 59.0 / 8, but 7.3 isn't).
Every floating point operation gives you the exact result, rounded to the nearest floating-point number. So you get
200.0 -> Exactly 200
273.15 -> A number very close to 273.15
200 - 273.15 -> A number very close to -73.15
-73.15 -> A number very close to -73.15
If you compare two numbers that are both very very close to -73.15 they are not necessarily equal. That's not a problem of the == operator; that one will determine correctly whether they are equal or not. The problem is that the two numbers can actually be different.