Swift: How to convert String to UInt?

问题

According to Swift - Converting String to Int, there's a String method toInt().

But, there's no toUInt() method. So, how to convert a String to a Uint?

回答1:

Update for Swift 2/Xcode 7:

As of Swift 2, all integer types have a (failable) constructor

init?(_ text: String, radix: Int = default)

which replaces the toInt() method of String, so no custom code is needed anymore for this task:

print(UInt("1234")) // Optional(1234)
// This is UInt.max on a 64-bit platform:
print(UInt("18446744073709551615")) // Optional(18446744073709551615)
print(UInt("18446744073709551616")) // nil (overflow)
print(UInt("1234x")) // nil (invalid character)
print(UInt("-12")) // nil (invalid character)

Old answer for Swift 1.x:

This looks a bit complicated, but should work for all numbers in the full range of UInt, and detect all possible errors correctly (such as overflow or trailing invalid characters):

extension String {
    func toUInt() -> UInt? {
        if contains(self, "-") {
            return nil
        }
        return self.withCString { cptr -> UInt? in
            var endPtr : UnsafeMutablePointer<Int8> = nil
            errno = 0
            let result = strtoul(cptr, &endPtr, 10)
            if errno != 0 || endPtr.memory != 0 {
                return nil
            } else {
                return result
            }
        }
    }
}

Remarks:

The BSD library function strtoul is used for the conversion. The endPtr is set to first "invalid character" in the input string, therefore endPtr.memory == 0 must be hold if all characters could be converted. In the case of a conversion error, the global errno variable is set to a non-zero value (e.g. ERANGE for an overflow).
The test for a minus sign is necessary because strtoul() accepts negative numbers (which are converted to the unsigned number with the same bit pattern).
A Swift string is converted to a C string "behind the scenes" when passed to a function taking a char * parameter, so one could be tempted to call strtoul(self, &endPtr, 0) (which is what I did in the first version of this answer). The problem is that the automatically created C string is only temporary and can already be invalid when strtoul() returns, so that endPtr does not point to a character in the input string anymore. This happened when I tested the code in the Playground. With self.withCString { ... }, this problem does not occur because the C string is valid throughout the execution of the closure.

Some tests:

println("1234".toUInt()) // Optional(1234)
// This is UInt.max on a 64-bit platform:
println("18446744073709551615".toUInt()) // Optional(18446744073709551615)
println("18446744073709551616".toUInt()) // nil (overflow)
println("1234x".toUInt()) // nil (invalid character)
println("-12".toUInt()) // nil (invalid character)

回答2:

You might be interested in a safer solution similar to:

let uIntString = "4"
let nonUIntString = "foo"

extension String {
    func toUInt() -> UInt? {
        let scanner = NSScanner(string: self)
        var u: UInt64 = 0
        if scanner.scanUnsignedLongLong(&u)  && scanner.atEnd {
            return UInt(u)
        }
        return nil
    }
}

uIntString.toUInt()     // Optional(4)
nonUIntString.toUInt()  // nil

Hope this helps

// Edited following @Martin R. suggestion

回答3:

Please, for the love of not crashing, don’t use ! to do this.

It’s easy to tack a map on the end of toInt to convert it to an optional UInt:

let str = "4"
let myUInt = str.toInt().flatMap { $0 < 0 ? nil : UInt($0) }

then use the usual unwrapping techniques on myUInt.

And if you find yourself doing this a lot:

extension String {
    func toUInt() -> UInt? {
        return self.toInt().flatMap { $0 < 0 ? nil : UInt($0) }
    }
}

let str = "-4"
if let myUInt = str.toUInt() {
    println("yay, \(myUInt)")
}
else {
    println("nuh-uh")
}

edit: as @MartinR points out, while safe, this doesn’t extract the full range of possible values for a UInt that Int doesn’t cover, see the other two answers.

回答4:

Use Forced Unwrapping or Optional Binding to make sure the string can be converted to UInt. eg:

let string = "123"
let number = UInt(string) //here number is of type *optional UInt*

//Forced Unwrapping

if number != nil {
  //converted to type UInt
}

回答5:

just use UInt's init:

let someString = "4"

UInt(someString.toInt()!) // 4

来源：https://stackoverflow.com/questions/30382414/swift-how-to-convert-string-to-uint

标签

string

swift

type-conversion

uint