我试图弄清楚如何在Swift中将Int转换为字符串。
我想出了一个解决方案，使用NSNumber，但我想弄清楚如何在Swift中做到这一切。
let x : Int = 45
let xNSNumber = x as NSNumber
let xString : String = xNSNumber.stringValue
当前回答
let intAsString = 45.description // "45"
let stringAsInt = Int("45") // 45
其他回答
将Unicode Int转换为字符串
对于那些想要将Int转换为Unicode字符串的人，您可以执行以下操作:
let myInteger: Int = 97
// convert Int to a valid UnicodeScalar
guard let myUnicodeScalar = UnicodeScalar(myInteger) else {
return ""
}
// convert UnicodeScalar to String
let myString = String(myUnicodeScalar)
// results
print(myString) // a
或者:
let myInteger: Int = 97
if let myUnicodeScalar = UnicodeScalar(myInteger) {
let myString = String(myUnicodeScalar)
}
在swift 3.0中，您可以将整数更改为字符串，如下所示
let a:String = String(stringInterpolationSegment: 15)
另一种方法是
let number: Int = 15
let _numberInStringFormate: String = String(number)
//或任何整数来代替15
斯威夫特4:
let x:Int = 45
let str:String = String(describing: x)
开发者。Apple.com >字符集init(describing:)
String(description:)初始化式是将任何类型的实例转换为字符串的首选方法。
自定义字符串转换
将String转换为Int
var numberA = Int("10")
Print(numberA) // It will print 10
将Int转换为String
var numberA = 10
1日)
print("numberA is \(numberA)") // It will print 10
2)
var strSomeNumber = String(numberA)
or
var strSomeNumber = "\(numberA)"
弦乐演奏
稍微讲讲性能 UI测试包在iPhone 7(真实设备)与iOS 14
let i = 0
lt result1 = String(i) //0.56s 5890kB
lt result2 = "\(i)" //0.624s 5900kB
lt result3 = i.description //0.758s 5890kB
import XCTest
class ConvertIntToStringTests: XCTestCase {
let count = 1_000_000
func measureFunction(_ block: () -> Void) {
let metrics: [XCTMetric] = [
XCTClockMetric(),
XCTMemoryMetric()
]
let measureOptions = XCTMeasureOptions.default
measureOptions.iterationCount = 5
measure(metrics: metrics, options: measureOptions) {
block()
}
}
func testIntToStringConstructor() {
var result = ""
measureFunction {
for i in 0...count {
result += String(i)
}
}
}
func testIntToStringInterpolation() {
var result = ""
measureFunction {
for i in 0...count {
result += "\(i)"
}
}
}
func testIntToStringDescription() {
var result = ""
measureFunction {
for i in 0...count {
result += i.description
}
}
}
}
