我更喜欢下面的实现，因为我不必担心，即使字符串是空的

let str = "abc"
str.popLast()

// Prints ab

str = ""
str.popLast() // It returns the Character? which is an optional

// Print <emptystring>

一个快速变化的类别:

extension String {
    mutating func removeCharsFromEnd(removeCount:Int)
    {
        let stringLength = count(self)
        let substringIndex = max(0, stringLength - removeCount)
        self = self.substringToIndex(advance(self.startIndex, substringIndex))
    }
}

使用:

var myString = "abcd"
myString.removeCharsFromEnd(2)
println(myString) // "ab"

全局的dropLast()函数作用于序列，因此也作用于字符串:

var expression  = "45+22"
expression = dropLast(expression)  // "45+2"

// in Swift 2.0 (according to cromanelli's comment below)
expression = String(expression.characters.dropLast())

斯威夫特4

var welcome = "Hello World!"
welcome = String(welcome[..<welcome.index(before:welcome.endIndex)])

or

welcome.remove(at: welcome.index(before: welcome.endIndex))

or

welcome = String(welcome.dropLast())

另一种方法如果你想从末尾删除一个或多个字符。

var myStr = "Hello World!"
myStr = (myStr as NSString).substringToIndex((myStr as NSString).length-XX)

其中XX是要删除的字符数。

斯威夫特4:

let choppedString = String(theString.dropLast())

在Swift 2中，这样做:

let choppedString = String(theString.characters.dropLast())

我推荐这个链接来了解Swift字符串。