swift - fatal error : Not enough bits to represent the passed value (Int16) in Swift

Question

我正在将一个库从 Java (Android) 转换为 -> Swift (iPhone)

Java 代码，工作正常:

   long a = 48590108397870l;
   short b = ((short)(a & 65535));//b == -28370

Swift 代码:

    let a : Int64  = 48590108397870
    let b: Int16 = Int16(a & 65535)//Fatal error: Not enough bits to represent the passed value

Answer 1

a & 65535 是介于 0 和 2¹⁶-1 之间的值，适合 UInt16，但不适合 Int16。与许多其他语言相反，Swift 不会隐式截断值。

整数有 init(truncatingIfNeeded:)做你想做的事的初始化器:

When the bit width of T (the type of source) is equal to or greater than this type’s bit width, the result is the truncated least-significant bits of source.

例子:

let a : Int64 = 48590108397870
let b = Int16(truncatingIfNeeded: a)
print(b) // -28370

另一种选择是先创建一个无符号整数，然后将其转换为具有相同位模式的有符号整数:

let a : Int64 = 48590108397870
let b = Int16(bitPattern: UInt16(a & 0xFFFF))
print(b) // -28370