我从我的服务器收到一个Int,我想爆炸到一个位掩码数组。因此,例如,如果我的服务器给我3号,我们得到两个值,二进制1和二进制2。
我如何在Swift中执行此操作?
答案 0 :(得分:4)
您可以使用:
let number = 3
//radix: 2 is binary, if you wanted hex you could do radix: 16
let str = String(number, radix: 2)
println(str)
打印“11”
let number = 79
//radix: 2 is binary, if you wanted hex you could do radix: 16
let str = String(number, radix: 16)
println(str)
打印“4f”
答案 1 :(得分:3)
我不知道任何好的内置方式,但你可以使用它:
var i = 3
let a = 0..<8
var b = a.map { Int(i & (1 << $0)) }
// b = [1, 2, 0, 0, 0, 0, 0, 0]
答案 2 :(得分:1)
这是一个简单的实现:
func intToMasks(var n: Int) -> [Int] {
var masks = [Int]()
var mask = 1
while n > 0 {
if n & mask > 0 {
masks.append(mask)
n -= mask
}
mask <<= 1
}
return masks
}
println(intToMasks(3)) // prints "[1,2]"
println(intToMasks(1000)) // prints "[8,32,64,128,256,512]"
答案 3 :(得分:0)
public extension UnsignedInteger {
/// The digits that make up this number.
/// - Parameter radix: The base the result will use.
func digits(radix: Self = 10) -> [Self] {
sequence(state: self) { quotient in
guard quotient > 0
else { return nil }
let division = quotient.quotientAndRemainder(dividingBy: radix)
quotient = division.quotient
return division.remainder
}
.reversed()
}
}
let digits = (6 as UInt).digits(radix: 0b10) // [1, 1, 0]
digits.reversed().enumerated().map { $1 << $0 } // [0, 2, 4]
如果需要,也可以反转结果。