Swift:如何将String转换为UInt?

问题描述

根据 Swift-将字符串转换为Int ，有一种String方法toInt().

但是，没有toUInt()方法.那么，如何将String转换为Uint?

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

推荐答案

Swift 2/Xcode 7更新:

从Swift 2开始，所有整数类型都有一个(失败的)构造函数

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

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

替换了String的toInt()方法，因此没有自定义此任务不再需要代码:

which replaces the toInt() method of String, so no customcode 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)

---

Swift 1.x的旧答案:

这看起来有点复杂，但是应该适用于UInt的完整范围，并正确检测所有可能的错误(例如溢出或尾随无效字符):

This looks a bit complicated, but should work for all numbers in thefull 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
            }
        }
    }
}

备注:

• BSD库函数 strtoul 用于转换.endPtr设置为输入字符串中的第一个无效字符"，因此，如果所有个字符，必须保留endPtr.memory == 0可以转换.如果发生转换错误，则设置全局errno变量设置为非零值(例如ERANGE表示溢出).

• 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 characterscould be converted.In the case of a conversion error, the global errno variable is setto a non-zero value (e.g. ERANGE for an overflow).

必须测试负号，因为strtoul()接受负数(使用负号转换为无符号数)相同的位模式).

The test for a minus sign is necessary because strtoul() acceptsnegative numbers (which are converted to the unsigned number with thesame bit pattern).

在以下情况下，Swift字符串将转换为幕后"的C字符串:传递给带有char *参数的函数，因此可能是试图调用strtoul(self, &endPtr, 0)(这是我在此答案的第一个版本).问题是自动创建的C字符串只是临时的，在以下情况下可能已经无效strtoul()返回，因此endPtr不会指向输入字符串中的字符不再存在.这是我在Playground中测试代码时发生的.使用self.withCString { ... }，不会发生此问题，因为C字符串在整个执行过程中都是有效的

A Swift string is converted to a C string "behind the scenes" whenpassed to a function taking a char * parameter, so one could betempted to call strtoul(self, &endPtr, 0) (which is what I did inthe first version of this answer). The problem is that the automaticallycreated C string is only temporary and can already be invalid whenstrtoul() returns, so that endPtr does not point to acharacter 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 executionof the closure.

一些测试:

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)

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