I just want to convert a character into an Int.
This should be simple. But I haven't found the previous answers helpful. There is always some error. Perhaps it is because I'm trying it in Swift 2.0.
for i in (unsolved.characters) {
fileLines += String(i).toInt()
print(i)
}
In Swift 2.0, toInt()
, etc., have been replaced with initializers. (In this case, Int(someString)
.)
Because not all strings can be converted to ints, this initializer is failable, which means it returns an optional int (Int?
) instead of just an Int
. The best thing to do is unwrap this optional using if let
.
I'm not sure exactly what you're going for, but this code works in Swift 2, and accomplishes what I think you're trying to do:
let unsolved = "123abc"
var fileLines = [Int]()
for i in unsolved.characters {
let someString = String(i)
if let someInt = Int(someString) {
fileLines += [someInt]
}
print(i)
}
Or, for a Swiftier solution:
let unsolved = "123abc"
let fileLines = unsolved.characters.filter({ Int(String($0)) != nil }).map({ Int(String($0))! })
// fileLines = [1, 2, 3]
You can shorten this more with flatMap
:
let fileLines = unsolved.characters.flatMap { Int(String($0)) }
flatMap
returns "an Array
containing the non-nil results of mapping transform
over self
"… so when Int(String($0))
is nil
, the result is discarded.
This is a bit of improvisational trick I came up with. Since it can be tricky to convert Character
to Int
, but you can easily convert String
to Int
do this :
fileLines = Int(String(i))!
of course this is not very well optimized but for cases like yours it can do the trick.
Actually. A simpler way is to convert String to Int in Swift 2.o is:
let chars = Int(chars)
Not sure if this is what you're trying for...But you can easily apply this to your loop, of course.
Swift 4 solutions
To convert a Character
to a String
:
let digit = Int(String("1"))!
//Don't force unwrap this if you're not 100% sure your character is a digit
Solution to OPs problem:
let fileLines = unsolved.compactMap{ Int(String($0) }
// fileLines = [1, 2, 3]
In the latest Swift versions (at least in Swift 5) you don't need to do additional conversion to String
.
Character
has property wholeNumberValue
which tries to convert a character to Int
and returns nil
if the character does not represent and integer.
let char: Character = "5"
if let intValue = char.wholeNumberValue {
print("Value is \(intValue)")
} else {
print("Not an integer")
}
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