I'm trying to detect the type of number a user enters into a textField. For example
if the user enters 1000 The program should return 1Kcs
if the user enters 12,000, The program should return 12Kcs
if the user enters 12,000,000 The program should return 12MCS
How do I go about this in swift?
Thousands - Kcs
Hundreds - Mcs
This should to the job
extension Int {
var unitFormatted: String {
let positive = self < 0 ? -self : self
switch positive {
case 1_000_000..<Int.max: return "\(self / 1_000_000)MCS"
case 1_000..<1_000_000: return "\(self / 1_000)Kcs"
default: return "\(self)"
}
}
}
Examples
0.unitFormatted // "0"
1.unitFormatted // "1"
1000.unitFormatted // "1Kcs"
12000.unitFormatted // "12Kcs"
12000000.unitFormatted // "12MCS"
Some variation of this to fit your needs should be fine:
var num = 12000000
switch num {
case 1000...999999:
print(String(num/1000) + "Kcs")
case 1000000...999999999:
print(String(num/1000000) + "Mcs")
default:
print(num)
}
Related
How do i get the result of of i to return from the function here? I can return it from inside the if { } but then I get an error there’s no global return (for the function). My goal is to write a function that accepts an Integer and returns the square root. My next step is to throw an error if it's not an Int or between 1 and 10,000, but please assume simple numbers for this question.
let userInputInteger = 9
func squareRootCheckpoint4(userInputInteger: Int) -> Int {
for i in 1...100 {
let userInputInteger = i * i
if i == userInputInteger {
break
}
}
return i
}
update:
To be able to return it, you will have to declared a variable outside the for-loop, then based on the logic assign the "i" counter to that outside variable.
func squareRootCheckpoint4(userInputInteger: Int) -> Int {
var result = 0 // this is to keep the result
for i in 1...100 {
let powerOfTwoResult = i * i
if powerOfTwoResult == userInputInteger {
result = i // assign result variable based on the logic
break
}
}
return result // return result
}
The error shown is because if the for-loop finishes without any hit on the if statement, the function will not be able to return any Int. That's why the compiler produces that "no global return" error.
One way to do it if you don't want to return a default value is to throw an error. You could throw an error from a function (https://docs.swift.org/swift-book/LanguageGuide/ErrorHandling.html). For example for your goal of returning square root of an integer and throw error if the input integer is not between 1 and 10000 (if I understand correctly), you could do something like this
enum InputError: Error {
case invalidInteger
}
func squareRootCheckpoint4(userInputInteger: Int) throws -> Int {
if userInputInteger < 1 || userInputInteger > 10000 {
throw InputError.invalidInteger
}
// calculate square root
return resultOfSquareroot
}
// and to handle the error, you could encapsulate the squareRootCheckpoint4 function in a do-catch statement
do {
let squareRootResult = try squareRootCheckpoint4(userInputInteger: 4)
} catch InputError.invalidInteger {
// handle your error here
}
Alternatively, you could do a validation on the input integer separately before calling the squareRootCheckpoint4 function. For example
func validate(_ input: Int) -> Bool {
if userInputInteger < 1 || userInputInteger > 10000 {
return false
}
return true
}
func squareRootCheckpoint4(userInputInteger: Int) -> Int {
// calculate square root
return resultOfSquareroot
}
var input: Int = 9
if validate(input) {
let squareRootResult = squareRootCheckpoint4(input)
} else {
// handle when input is invalid
}
A couple of things are swapped around, and something (anything) needs to be returned from the function. The framing of the question only allows solutions that don't work if the input is not an integer with a square root.
List of Specifics:
The function argument name userInputInteger should not have been used to instantiate a new object. That's confusing. I did it because I'm learning and thought it was necessary.
Instead, i times i (or i squared) should have been assigned to a new object.
The if statement should be equal to this new object, instead of i.
Functions should return something. When i is returned in the if statement, it's not available anywhere outside the for loop. The print statements in the code example help explain what is available in these different scopes, as well as what's not.
The for loop stops when i is returned. That happens when the guess (i) is equal to the input (userInputInteger).
The for loop will continue through 100 if the input doesn't have a square root.
The function should return the correct number that the if statement was trying to guess, but you have to tell it to (with a return statement) and do it in the right spot. This is the "global return" error. However, because this approach in the question is setup poorly without error handling, only an Integer with a sqare root will return correctly. And in this case, the function's global return doesn't matter; Swift just requires something be returned, regardless of whether or not it's used.
Code Example - Direct Answer:
import Cocoa
let userInputInteger = 25
let doesNotMatter = 0
print("sqrt(\(userInputInteger)) is \(sqrt(25))") //correct answer for reference
func squareRootCheckpoint4(userInputInteger2: Int) -> Int {
var j: Int
for i in 1...100 {
print("Starting Loop \(i)")
j = i * i
if j == userInputInteger2 {
print("i if-loop \(i)")
print("j if-loop \(j)")
return i
}
print("i for-loop \(i)")
print("j for-loop \(j)")
}
//print("i func-end \(i)") //not in scope
//print("j func-end \(j)") //not in scope
//return i //not in scope
//return j //not in scope
print("Nothing prints here")
// but something must be returned here
return doesNotMatter
}
print("Input was \(userInputInteger)")
print("The Square Root of \(userInputInteger) is \(squareRootCheckpoint4(userInputInteger2: userInputInteger))")
Code Example - Improved with Error Handling
import Cocoa
enum ErrorMsg : Error {
case outOfBoundsError, noRootError
}
func checkSquareRoot (Input userInputInteger2: Int) throws -> Int {
if userInputInteger < 1 || userInputInteger > 10_000 {
throw ErrorMsg.outOfBoundsError
}
var j : Int
for i in 1...100 {
print("Starting Loop \(i)")
j = i * i
if j == userInputInteger2 {
print("i if-loop \(i)")
print("j if-loop \(j)")
return i
}
print("i for-loop \(i)")
print("j for-loop \(j)")
}
throw ErrorMsg.noRootError
}
let userInputInteger = 25
print("Input was \(userInputInteger)")
do {
let result = try checkSquareRoot(Input: userInputInteger)
print("The Square Root of \(userInputInteger) is \(result)")
} catch ErrorMsg.outOfBoundsError {
print("out of bounds: the userInputInteger is less than 1 or greater than 10,000")
} catch ErrorMsg.noRootError {
print("no root")
}
print("Swift built-in function for reference: sqrt(\(userInputInteger)) is \(sqrt(25))") //correct answer for reference
while true {
print ( """
1. Log in
2. Create new user
3. Quit
""")
if let numberString = readLine(), let number = Int(numberString) {
print("you entered \(number)")
break
} else {
print("Try again")
}
}
It is necessary to use the number variable in the future to compare it.
How can it be taken out of sight?
When I try to make a global var something goes wrong:
var numberString: String?
var number: Int?
while true {
print ( """
1. Log in
2. Create new user
3. Quit
""")
if numberString = readLine(), number = Int(numberString) {
print("you entered \(number)")
break
} else {
print("Try again")
}
}
Considering that you are creating a command prompt and that choice has no meaning outside your loop there is no need to make it global. You only need to switch the user selection and decide what to do from there. Note that if you try to break from inside the switch you won't exit your loop as I showed to you in your last question. To allow the compiler to know that you want to break the while loop instead of the switch you need to label your while loop statement, this way you can specify what you want to break when you have a switch inside a loop. Try like this:
func getValue() -> Int? {
guard let line = readLine(), let value = Int(line) else {
return nil
}
return value
}
question: while true {
print("""
1. Log in
2. Create new user
3. Quit
""")
guard let value = getValue() else {
continue
}
switch value {
case 1:
print("you have selected number one")
case 2:
print("you have selected number two")
case 3:
print("Good bye")
break question
default:
print("Try again")
}
}
There is not much information to go off of here, but I think I can help. In the first script, you are creating a new variable and setting the value for that variable:
if let numberString = readLine(), let number = Int(numberString) { ... }
In your second script, you have to global variables, but at no point have you provided them a value. What you need to do is provide both global variables (numberString and number) a value before comparing them. As such, you need to use the == sign to compare different variables/ types. For instance:
var numberString: String?
var number: Int?
while true {
print("""
1. Log in
2. Create new user
3. Quit
""")
numberString = readLine()
number = Int(numberString!)
}
You can now compare and or print your global variables...
Final Code
var numberString: String?
var number: Int?
while true {
print("""
1. Log in
2. Create new user
3. Quit
""")
numberString = readLine()
number = Int(numberString!)
if number == 1 {
print("you have selected number one")
} else if number == 2 {
print("you have selected number two")
} else if number == 3 {
print("you have selected number three")
} else {
print("Try again")
}
}
Due to the fact that both variables numberString and number are now global variables, you can access them in the future if you need to compare them again.
I am trying to code a function that can give me a message when two int numbers get close.
Like:
static let too_low_number = "The number is too low."
static let low_number = "The number is low."
static let too_high_number = "The number is too high."
static let high_number = "The number is high."
var closemove = 10
var appanswer = Int.random(in: 1...100)
var playeranswer = 0
if playeranswer gets 10 int close to appanswer then it gives a message. Get 10 int close but less-than appanswer then it gives me low_number and Get bigger distance than 10 int but less-than appanswer then it gives me too_low_number.
same as getting 10 int close but greater-than appanswer then it gives me high_number and Get bigger distance than 10 int but greater-than appanswer then it gives me too_high_number.
You can check the difference between the player and the correct answer using switch with intervals for this. Here is a solution using switch in a computed property.
Please follow swift naming conventions of variables by using snake case, I took the liberty to rename all the variables in my code
var result: String {
switch playerAnswer - appAnswer {
case -closeMove..<0:
return lowNumber
case ..<(-closeMove):
return tooLowNumber
case 1...closeMove:
return highNumber
case (closeMove + 1)...:
return tooHighNumber
default:
return "Right answer"
}
}
I code a function:
func result()-> String{
var takeEx = 0
if (getplayer < setApp) {
takeEx = setApp - getplayer
if (takeEx < 10) {
return low_number
} else {
return too_low_number
}
} else if (getplayer > setApp){
takeEx = getplayer - setApp
if (takeEx < 10) {
return high_number
} else {
return too_high_number
}
} else if (getplayer == setApp){
return congratulation
}
}
}
I have this piece of code
func sell() throws{
while(true)
{
var choice : String?
print("Please press a number from 1 to 3\n")
let product = readLine(stripNewline: true)!
switch product
{
case "1":
//
case "2":
//
case "3":
//
default:
choice = "Invalid"
try sell()
}
}
try sell()
And it gives me the error
execution was interrupted reason exc_bad_instruction
I realized that the ! is causing the error. If I remove it I have a problem with the comparisons inside switch.
Anyone knows what is the problem?
func foo()->Int {
print("Please press a number from 1 to 3")
while true {
if let l = readLine(),
let i = Int(l) {
if (1..<4).contains(i) {
return i
} else {
print("Number must be in range from 1 to 3")
}
} else {
print("Please press a number")
}
}
}
let r = foo()
print("You chose", r)
an example of ...
Please press a number from 1 to 3
u
Please press a number
7
Number must be in range from 1 to 3
45
Number must be in range from 1 to 3
h76
Please press a number
1
You chose 1
Program ended with exit code: 0
How about create one if let to check if this contains value or is it nil
let product = readLine(stripNewline: true)
if let productNo = product {
switch productNo {
//your same code in switch
}
I'm trying to change a tuple in an array , however ,when I try
emo = (type:emo.type,strength:increaseStrength(emo.strength))
it gives me error
"cannot assign to 'let' value 'emo'
here is my code :
var emotions : [(type : String, strength: Int)] = [("happy",0),("scared",0),("tender",0),("excited",0),("sad",0)]
func increaseStrength(i:Int)->Int {
switch i {
case 0: return 1
case 1: return 2
case 2: return 3
case 3: return 0
default :return 0
}
}
#IBAction func HappyBA(sender: AnyObject) {
for emo in emotions {
if (emo.type == "happy" ){
emo = (type:emo.type,strength:increaseStrength(emo.strength))
}
}
println(emotions)
}
If there are better way to do the assignment please tell me I am so appreciated ! Thanks..
There is no point assigning to emo even if you could do it. This is not the same as replacing the corresponding object in the array - which is what you do want to do. emo is a copy; even if you were to set a property of it, it wouldn't affect the one back in the array. And certainly setting the variable would not magically read back into the array!
Here's one solution. Instead of cycling thru emotions in your for-loop, cycle thru enumerate(emotions). Now you have a tuple of an index number along with an emotion. If this is the right emotion type, write into the array via the index number.
for (ix,emo) in enumerate(emotions) {
if emo.type == "happy" {
emotions[ix] = (type:emo.type,strength:increaseStrength(emo.strength))
}
}
Or you could use map.
emotions = emotions.map {
emo in
if emo.type == "happy" {
return (type:emo.type,strength:increaseStrength(emo.strength))
} else {
return emo
}
}