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如何快速将double转换为字节数组?

面试问答 更新时间: 发布时间: IT归档 最新发布 模块sitemap 名妆网 法律咨询 聚返吧 英语巴士网 伯小乐 网商动力

如何快速将double转换为字节数组?

typealias Byte = UInt8

func toByteArray<T>(var value: T) -> [Byte] {    return withUnsafePointer(&value) {        Array(UnsafeBufferPointer(start: UnsafePointer<Byte>($0), count: sizeof(T)))    }}toByteArray(1729.1729)toByteArray(1729.1729 as Float)toByteArray(1729)toByteArray(-1729)

但是结果与您的期望相反(由于字节顺序):

[234, 149, 178, 12, 177, 4, 155, 64][136, 37, 216, 68][193, 6, 0, 0, 0, 0, 0, 0][63, 249, 255, 255, 255, 255, 255, 255]

添加:

func fromByteArray<T>(value: [Byte], _: T.Type) -> T {    return value.withUnsafeBufferPointer {        return UnsafePointer<T>($0.baseAddress).memory    }}let a: Double = 1729.1729let b = toByteArray(a) // -> [234, 149, 178, 12, 177, 4, 155, 64]let c = fromByteArray(b, Double.self) // -> 1729.1729

对于Xpre8 / Swift3.0:

func toByteArray<T>(_ value: T) -> [UInt8] {    var value = value    return withUnsafePointer(to: &value) {        $0.withMemoryRebound(to: UInt8.self, capacity: MemoryLayout<T>.size) { Array(UnsafeBufferPointer(start: $0, count: MemoryLayout<T>.size))        }    }}func fromByteArray<T>(_ value: [UInt8], _: T.Type) -> T {    return value.withUnsafeBufferPointer {        $0.baseAddress!.withMemoryRebound(to: T.self, capacity: 1) { $0.pointee        }    }}

对于Xpre8.1 / Swift3.0.1

func toByteArray<T>(_ value: T) -> [UInt8] {    var value = value    return withUnsafeBytes(of: &value) { Array($0) }}func fromByteArray<T>(_ value: [UInt8], _: T.Type) -> T {    return value.withUnsafeBytes {        $0.baseAddress!.load(as: T.self)    }}


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