Does anybody know if there is a way of removing (trimming) the last two digits or first two digits from a number. I have the number 4131 and I want to separate that into 41 and 31, but after searching the Internet, I've only managed to find how to remove characters from a string, not a number. I tried converting my number to a string, then removing characters, and then converting it back to a number, but I keep receiving errors.
I believe I will be able to receive the first two digit by dividing the number by 100 and then rounding the number down, but I don't have an idea of how to get the last two digits?
Does anybody know the function to use to achieve what I'm trying to do, or can anybody point me in the right direction.
Try this:
var num = 1234
var first = num/100
var last = num%100
The playground's output is what you need.
You can use below methods to find the last two digits
func getLatTwoDigits(number : Int) -> Int {
return number%100; //4131%100 = 31
}
func getFirstTwoDigits(number : Int) -> Int {
return number/100; //4131/100 = 41
}
To find the first two digit you might need to change the logic on the basis of face value of number. Below method is generalise method to find each digit of a number.
func printDigits(number : Int) {
var num = number
while num > 0 {
var digit = num % 10 //get the last digit
println("\(digit)")
num = num / 10 //remove the last digit
}
}
Related
I am building an app that the user selects a multiplication table. Then it gives random numbers to multiplicate with the number they select. for example, if the user selects "1". the questions shown would be "1 x 1", or "1 x 8".
The problem is that I need to assign the same random number to 2 variables. The one that will be shown on the question and the one used to calculate the result.
I thought of something like this, but the random number is different on each variable. What can I do to use the same random number generated on 2 variables?
func game() -> (String, Int) {
let randomNumber = multiplicate.randomElement()
switch selectedQuestion {
case 0:
return ("1 × \(randomNumber!)", 1 * randomNumber!)
default:
return ("", 0)
}
}
Not exactly sure what you're asking but you could do something like this:
let number = Int.random(in: 0..<10)
let number2 = number
but I don't think there is a need to create a new variable for this. The whole idea of variables is that you can save some value and then use it later so there isn't really a need to create number2 here.
I want to format, in real time, the number entered into a UITextField. Depending on the field, the number may be an integer or a double, may be positive or negative.
Integers are easy (see below).
Doubles should be displayed exactly as the user enters with three possible exceptions:
If the user begins with a decimal separator, or a negative sign followed by a decimal separator, insert a leading zero:
"." becomes "0."
"-." becomes "-0."
Remove any "excess" leading zeros if the user deletes a decimal point:
If the number is "0.00023" and the decimal point is deleted, the number should become "23".
Do not allow a leading zero if the next character is not a decimal separator:
"03" becomes "3".
Long story short, one and only one leading zero, no trailing zeros.
It seemed like the easiest idea was to convert the (already validated) string to a number then use format specifiers. I've scoured:
https://developer.apple.com/library/content/documentation/Cocoa/Conceptual/Strings/Articles/formatSpecifiers.html
and
http://www.cplusplus.com/reference/cstdio/printf/
and others but can't figure out how to format a double so that it does not add a decimal when there are no digits after it, or any trailing zeros. For example:
x = 23.0
print (String(format: "%f", x))
//output is 23.000000
//I want 23
x = 23.45
print (String(format: "%f", x))
//output is 23.450000
//I want 23.45
On How to create a string with format?, I found this gem:
var str = "\(INT_VALUE) , \(FLOAT_VALUE) , \(DOUBLE_VALUE), \(STRING_VALUE)"
print(str)
It works perfectly for integers (why I said integers are easy above), but for doubles it appends a ".0" onto the first character the user enters. (It does work perfectly in Playground, but not my program (why???).
Will I have to resort to counting the number of digits before and after the decimal separator and inserting them into a format specifier? (And if so, how do I count those? I know how to create the format specifier.) Or is there a really simple way or a quick fix to use that one-liner above?
Thanks!
Turned out to be simple without using NumberFormatter (which I'm not so sure would really have accomplished what I want without a LOT more work).
let decimalSeparator = NSLocale.current.decimalSeparator! as String
var tempStr: String = textField.text
var i: Int = tempStr.count
//remove leading zeros for positive numbers (integer or real)
if i > 1 {
while (tempStr[0] == "0" && tempStr[1] != decimalSeparator[0] ) {
tempStr.remove(at: tempStr.startIndex)
i = i - 1
if i < 2 {
break
}
}
}
//remove leading zeros for negative numbers (integer or real)
if i > 2 {
while (tempStr[0] == "-" && tempStr[1] == "0") && tempStr[2] != decimalSeparator[0] {
tempStr.remove(at: tempStr.index(tempStr.startIndex, offsetBy: 1))
i = i - 1
if i < 3 {
break
}
}
}
Using the following extension to subscript the string:
extension String {
subscript (i: Int) -> Character {
return self[index(startIndex, offsetBy: i)]
}
}
I am new in Swift.
I am trying to make a budget application. This app have a Calculator like keyboard. My idea is when users enter the money app will automatically add a decimal place for users.
For example, if you type 1230 it will give you 12.30 and type 123 it will display 1.23
I wrote a couple lines of code down below. The problem is it only can add decimal point after first digit it won't go backwards when you give more digits. It only can display as X.XXXXX
I tried solve this problem with String.index(maybe increase index?) and NSNumber/NSString format. But I don't know this is the right direction or not.
let number = sender.currentTitle!
let i: String = displayPayment.text!
if (displayPayment.text?.contains("."))!{
displayPayment.text = i == "0" ? number : displayPayment.text! + number
}
else {
displayPayment.text = i == "0" ? number : displayPayment.text! + "." + number
}
Indexing Strings in Swift is not as "straightforward" as many would like, simply due to how Strings are represented internally. If you just want to add a . at before the second to last position of the user input you could do it like this:
let amount = "1230"
var result = amount
if amount.characters.count >= 2 {
let index = amount.index(amount.endIndex, offsetBy: -2)
result = amount[amount.startIndex..<index] + "." + amount[index..<amount.endIndex]
} else {
result = "0.0\(amount)"
}
So for the input of 1230 result will be 12.30. Now You might want to adjust this depending on your specific needs. For example, if the user inputs 30 this code would result in .30 (this might or might not be what you want).
Let's say I want to generate a random number between 1 and 100, but I don't want to include 42. How would I do this without repeating the random method until it is not 42.
Updated for Swift 5.1
Excluding 1 value
var nums = [Int](1...100)
nums.remove(at: 42)
let random = Int(arc4random_uniform(UInt32(nums.count)))
print(nums[random])
Excluding multiple values
This extension of Range does provide a solution when you want to exclude more than 1 value.
extension ClosedRange where Element: Hashable {
func random(without excluded:[Element]) -> Element {
let valid = Set(self).subtracting(Set(excluded))
let random = Int(arc4random_uniform(UInt32(valid.count)))
return Array(valid)[random]
}
}
Example
(1...100).random(without: [40,50,60])
I believe the computation complexity of this second solution is O(n) where n is the number of elements included in the range.
The assumption here is the no more than n excluded values are provided by the caller.
appzYourLife has some great general purpose solutions, but I want to tackle the specific problem in a lightweight way.
Both of these approaches work roughly the same way: Narrow the range to the random number generator to remove the impossible answer (99 answers instead of 100), then map the result so it isn't the illegal value.
Neither approach increases the probability of an outcome relative to another outcome. That is, assuming your random number function is perfectly random the result will still be random (and no 2x chance of 43 relative to 5, for instance).
Approach 1: Addition.
Get a random number from 1 to 99. If it's greater than or equal to the number you want to avoid, add one to it.
func approach1()->Int {
var number = Int(arc4random_uniform(99)+1)
if number >= 42 {
number = number + 1
}
return number
}
As an example, trying to generate a random number from 1-5 that's not 3, take a random number from 1 to 4 and add one if it's greater than or equal to 3.
rand(1..4) produces 1, +0, = 1
rand(1..4) produces 2, +0, = 2
rand(1..4) produces 3, +1, = 4
rand(1..4) produces 4, +1, = 5
Approach 2: Avoidance.
Another simple way would be to get a number from 1 to 99. If it's exactly equal to the number you're trying to avoid, make it 100 instead.
func approach2()->Int {
var number = Int(arc4random_uniform(99)+1)
if number == 42 {
number = 100
}
return number
}
Using this algorithm and narrowing the range to 1-5 (while avoiding 3) again, we get these possible outcomes:
rand(1..4) produces 1; allowed, so Result = 1
rand(1..4) produces 2, allowed, so Result = 2
rand(1..4) produces 3; not allowed, so Result = 5
rand(1..4) produces 4, allowed, so Result = 4
I want to convert the decimal number 27 into binary such a way that , first the digit 2 is converted and its binary value is placed in an array and then the digit 7 is converted and its binary number is placed in that array. what should I do?
thanks in advance
That's called binary-coded decimal. It's easiest to work right-to-left. Take the value modulo 10 (% operator in C/C++/ObjC) and put it in the array. Then integer-divide the value by 10 (/ operator in C/C++/ObjC). Continue until your value is zero. Then reverse the array if you need most-significant digit first.
If I understand your question correctly, you want to go from 27 to an array that looks like {0010, 0111}.
If you understand how base systems work (specifically the decimal system), this should be simple.
First, you find the remainder of your number when divided by 10. Your number 27 in this case would result with 7.
Then you integer divide your number by 10 and store it back in that variable. Your number 27 would result in 2.
How many times do you do this?
You do this until you have no more digits.
How many digits can you have?
Well, if you think about the number 100, it has 3 digits because the number needs to remember that one 10^2 exists in the number. On the other hand, 99 does not.
The answer to the previous question is 1 + floor of Log base 10 of the input number.
Log of 100 is 2, plus 1 is 3, which equals number of digits.
Log of 99 is a little less than 2, but flooring it is 1, plus 1 is 2.
In java it is like this:
int input = 27;
int number = 0;
int numDigits = Math.floor(Log(10, input)) + 1;
int[] digitArray = new int [numDigits];
for (int i = 0; i < numDigits; i++) {
number = input % 10;
digitArray[numDigits - i - 1] = number;
input = input / 10;
}
return digitArray;
Java doesn't have a Log function that is portable for any base (it has it for base e), but it is trivial to make a function for it.
double Log( double base, double value ) {
return Math.log(value)/Math.log(base);
}
Good luck.