Converting Int to Bool

33
votes

In Swift 2.x I believe I could do:

let number = 1
let result = Bool(number)

print(result) // prints out: true

But since Swift 3 I've been unable to do this and it gives me the error:

Cannot invoke initialiser for type 'Bool' with an argument list of type '(Int)'

Currently I'm using an extension to convert an Int to a Bool but I was wondering if there isn't a build in option to do this.

7
swiftintbooleanswift3
It's not entirely clear what "converting an int to a bool" even means. Historically, languages like C treat non-zero as true, and 0 as false, but that's just a consequence of the implementation details of encoding true vs false in memory. When you think about it, it actually makes little sense, semantically. So instead, just use explicitly i != 0, to be clear about precisely what you mean. Over-assuming its generalizability and naming it something like boolValue is ambiguous. For one, Bash/linux exit codes are the opposite. 0 is good/true, non-zero is bad/false. - Alexander

7 Answers

60
votes

No, there is and has never been an explicit built in option for conversion of Int to Bool, see the language reference for Bool for details.

There exists, still, however, an initializer by NSNumber. The difference is that implicit bridging between Swift numeric type and NSNumber has been removed in Swift 3 (which previously allowed what seemed to be explicit Bool by Int initialization). You could still access this by NSNumber initializer by explicitly performing the conversion from Int to NSNumber:

let number = 1
let result = Bool(number as NSNumber)

print(result) // true

As @Hamish writes in his comment below: if we leave the subject of initializers and just focus on the end result (instantiating a Bool instance given the value of an Int instance) we can simply make use of the != operator for Int values (specifically, the operator with signature func !=(lhs: Int, rhs: Int) -> Bool), a generalization easily achievable using the != operator approach:

let number = -1
let result = number != 0

print(result) // true

Much like you yourself as well as @JAL describes in his answer, you could construct your own Bool by Int initializer, but you might as well consider generalizing this for any type conforming to the Integer protocol:

extension Bool {
    init<T: Integer>(_ num: T) {
        self.init(num != 0)
    }
}

/* example usage */
let num1: Int8 = -1
let num2: Int = 3
let num3: UInt64 = 0
// ....
let result1 = Bool(num1) // true
let result2 = Bool(num2) // true
let result3 = Bool(num3) // false
30
votes

Swift 4

extension Int {
    var boolValue: Bool { return self != 0 }
}

Swift 3

extension Integer {
    var boolValue: Bool { return self != 0 }
}

Usage

let number = 2
print(number.boolValue)

let items = ["1"]
print(items.count.boolValue)
6
votes

There is no Boolean initializer that takes an Int, only NSNumber. Previously, the Int was implicitly bridged to NSNumber through Foundation, but that was removed in Swift 3.

You can do something like:

let result = Bool(number as NSNumber)

Or, you can extend Bool and create a custom init that takes an Int:

extension Bool {    
    init(_ number: Int) {
        self.init(number as NSNumber)
    }
}
6
votes

I'm using Xcode 9.0.1 and Swift 3.0. 

let result = (number as NSNumber).boolValue

which is working very well for me.

2
votes

Swift 5

let number = 1
let result = Bool(truncating: number as NSNumber) //here result will be false if the number's value is 0 and it will be true for any other number's value

print(result)
1
votes
let iFalse: Int = 0
let iTrue: Int = 1
let iNil: Int = 2 //or any number other than 0 & 1

Bool(exactly: iFalse as NSNumber) //false
Bool(exactly: iTrue as NSNumber) //true
Bool(exactly: iNil as NSNumber) //nil
0
votes

Swift 5 try this 

extension Int32 {
    var boolValue: Bool {
        return (self as NSNumber).boolValue
    }
}

or

extension Int32 {
    var boolValue: Bool {
        return Bool.init(truncating: NSNumber.init(value: self))
    }
}