奇怪的 Swift 数字类型转换

问题描述

我刚刚注意到 Swift 对 Int 和 Double 进行了一些类型转换.当我尝试评估

I've just noticed that Swift does some type casting over Int and Double.When I try to evaluate

(10 / 3.0) - (10 / 3)

0.333... 是预期的，但实际上是 0.0.有人可以解释一下吗?

0.333... is expected, but it's actually 0.0.Could someone explain this please?

推荐答案

是的，我也觉得这很令人惊讶.Double 符合 FloatLiteralConvertible 和 IntegerLiteralConvertible(Swift 3 中的 ExpressibleByFloatLiteral 和 ExpressibleByIntegerLiteral).因此一个Double 可以用浮点字面量

Yes, I also found this quite surprising. Double conforms to both FloatLiteralConvertible and IntegerLiteralConvertible (ExpressibleByFloatLiteral and ExpressibleByIntegerLiteral in Swift 3). Therefore aDouble can be initialized with floating point literal

let a = 3.0

或使用整数文字:

let b : Double = 10

(对于其他浮点类型，如 Float 和CGFloat.)

(The same is true for other floating point types like Float andCGFloat.)

现在对我们所有有(Objective-)C背景的人来说可能是出乎意料的这两个陈述

Now it might be unexpected for all of us with an (Objective-)C backgroundthat both statements

let x : Double = 10/4     // x = 2.5 .  Really? Yes!
let y = 10/4 as Double    // Same here ...

将值 0.25 分配给变量.从上下文来看，结果除法必须是 Double 并且 Swift 不会隐式转换类型.因此/ 必须是浮点除法运算符

assign the value 0.25 to the variable. From the context, the result of thedivision must be a Double and Swift does not implicitly convert types.Therefore / must be the floating point division operator

func /(lhs: Double, rhs: Double) -> Double

所以编译器从文字中创建两个参数作为 Doubles10"和4".(如果 10/4 被视为两个整数的除法那么结果也将是一个整数，并且不能被赋值到 Double.)

so the compiler creates both arguments as Doubles from the literals"10" and "4". (If 10/4 were treated as the division of two integersthen the result would also be an integer, and that cannot be assignedto a Double.)

注意这与

let z = Double(10/4)   // z = 2.0 . (I just thought that I understood it &%$!?)

进行整数除法并将结果转换为Double.Double 有一个 init(_ v: Int) 构造函数，因此 10/4可以在这里被视为两个整数的除法.

which does an integer division and converts the result to Double.Double has an init(_ v: Int) constructor, and therefore 10/4can be treated as the division of two integers here.

如果我们总结这些结果，真的有点奇怪:

It really looks a bit strange if we summarize these results:

let x : Double = 10/4     // x = 2.5
let y = 10/4 as Double    // y = 2.5
let z = Double(10/4)      // z = 2.0

现在我们可以将这些结果应用到您的表达式中

Now we can apply these results to your expression

(10 / 3.0) - (10 / 3)

第一部分(10/3.0)只能是一个Double，因此-必须是浮点减法运算符

The first part (10 / 3.0) can only be a Double, therefore -must be the floating point subtraction operator

func -(lhs: Double, rhs: Double) -> Double

因此 (10/3) 也必须是 Double.同样，/ 必须是浮点除法运算符，所以 10 和 3 被视为 Double 常量.

and thus (10 / 3) must also be a Double. Again, / must be the floating point division operator, so 10 and 3 are treated as Double constants.

因此表达式等价于

(Double(10) / 3.0) - (Double(10) / Double(3))

并计算为 0.0.如果将表达式更改为

and evaluates to 0.0. If you change the expression to

(10 / 3.0) - Double(10 / 3)

那么结果是 0.333... 因为在这种情况下，10/3是两个整数常量的除法，如上所述.

then the result is 0.333... because in this context, 10 / 3is the division of two integer constants, as explained above.

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