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

我正在使用这个简单的方法

字符串到Int:

 var a = Int()
var string1 = String("1")
a = string1.toInt()

和从Int到String:

var a = Int()
a = 1
var string1 = String()
 string1= "\(a)"

其他回答

斯威夫特2:

var num1 = 4
var numString = "56"
var sum2 = String(num1) + numString
var sum3 = Int(numString)

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

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
            }
        }
    }
}

如果你喜欢快速扩展,你可以添加以下代码

extension Int
{
    var string:String {
        get {
            return String(self)
        }
    }
}

然后,您可以通过刚才添加的方法获取字符串

var x = 1234
var s = x.string
let a =123456888
var str = String(a)

OR

var str = a as! String