deadbeef - 4 months ago 13
Swift Question

Failable Float to Int conversion in Swift

I'm trying to find a reliable way to convert a

`Float`
or a
`Double`
to an
`Int`
in Swift. I run into issues when there is an overflow.

Let's take the following example:

``````let double = 9223372036854775807.0 // This is 2^63 - 1 (aka Int.max on 64 bits architecture)

print("Int.max is         : \(Int.max)")
print(String(format: "double value is    : %f", double))
print(String(format: "Double(Int.max) is : %f", Double(Int.max)))

let int: Int
if (double >= Double(Int.max)) {
print("Warning : double is too big for int !")
int = Int.max
} else {
int = Int(double)
}
``````

Which prints:

`````` Int.max is         : 9223372036854775807
double value is    : 9223372036854775808.000000
Double(Int.max) is : 9223372036854775808.000000
Warning : double is too big for int !
``````

This approach is very cumbersome. Besides, you probably noticed that I test for
`double >= Double(Int.max)`
and not
`double > Double(Int.max)`
. That's because
`Double(Int.max)`
is actually equal to
`Int.max + 1`
because of rounding. And my code should probably have to be different for 32 bits architectures.

So is there another way? Like a failable initializer that I would have missed or a better, portable way to do this?

You can extend `Int` by creating your own failable initializer for `Double`:

``````extension Int {
init?(double: Double) {
if double >= Double(Int.max) || double < Double(Int.min) || double.isNaN || double.isInfinite {
return nil
} else {
self = Int(double)
}
}
}

if let i = Int(double: 17) {
print(i)
} else {
print("nil")
}

// It also handles NaN and Infinite
let nan = sqrt(-1.0)
let inf = 1e1000

Int(double: nan)   // nil
Int(double: inf)   // nil
``````
Source (Stackoverflow)