奇怪的Swift数字类型转换

问题描述

我刚刚注意到Swift做了一些类型转换Int和Double。
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 0.0 。
有人可以解释这个吗？

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

推荐答案

是的，我也发现这很奇怪。 Double 符合 FloatLiteralConvertible 和 IntegerLiteralConvertible （ ExpressibleByFloatLiteral 和 ExpressibleByIntegerLiteral ）。因此，
Double 可以使用浮点文本初始化

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

/ em>：

let b : Double = 10

（对于其他浮点类型也是如此，例如 Float 和
CGFloat

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

现在，对于所有拥有（Objective-）C后台的人来说，这可能是意外的。
这两个语句





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 ...

值 0.25 添加到变量。从上下文中，
除法的结果必须是 Double ，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

因此编译器创建两个参数 Double s从字面值
10和4。 （如果 10/4 被视为两个整数
的除法，则结果也将是一个整数，不能将
赋值给 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.)

请注意，这不同于

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

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。

Now we can apply these results to your expression

(10 / 3.0) - (10 / 3)

第一部分（10 / 3.0）只能是 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，因此 10 和 3 被视为 Double 常数。

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，表达式等效于

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

并计算 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 ... ，因为在此上下文中， 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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