对于1字节的Int8,它返回两个字母,例如0A

对于2字节的Int16,它返回四个字母,例如0A0B

对于8位Int64,它返回16个字母,例如0102030405060708

func hex(v: Int) -> String {
    var s = ""
    var i = v
    for _ in 0..<sizeof(Int)*2 {
        s = String(format: "%x",i & 0xF) + s
        i = i >> 4
    }
    return s
}

func hex(v: Int64) -> String {
    var s = ""
    var i = v
    for _ in 0..<sizeof(Int64)*2 {
        s = String(format: "%x",i & 0xF) + s
        i = i >> 4
    }
    return s
}

func hex(v: Int32) -> String {
    var s = ""
    var i = v
    for _ in 0..<sizeof(Int32)*2 {
        s = String(format: "%x",i & 0xF) + s
        i = i >> 4
    }
    return s
}

func hex(v: Int16) -> String {
    var s = ""
    var i = v
    for _ in 0..<sizeof(Int16)*2 {
        s = String(format: "%x",i & 0xF) + s
        i = i >> 4
    }
    return s
}

func hex(v: Int8) -> String {
    var s = ""
    var i = v
    for _ in 0..<sizeof(Int8)*2 {
        s = String(format: "%x",i & 0xF) + s
        i = i >> 4
    }
    return s
}

上面的代码工作正常.

然后我尝试创建一个通用版本,如下所示：

func hex<T: IntegerType>(v: T) -> String {
    var s = ""
    var i = v
    for _ in 0..<sizeof(T)*2 {
        s = String(format: "%x",i & 0xF) + s
        i = i >> 4
    }
    return s
}

编译此代码时,我收到错误：T不能转换为Int

什么是正确的方式来实现这个任务？