Question

我认为,用一个简单的例子来解释这一点比较容易。 (帮助重写标题是受欢迎的;-)

我愿采用<条码>,使用<条码>,并使用<条码>简单明了的代号/代码>,自动将其添加到支持<条码>的任何类别中。

它非常容易:

class EnhancedInt(x: Int) { def squared = x * x }

implicit def IntToEnchancedInt(x: Int) = new EnhancedInt(x)

但是,如果存在任何或任何错误,我会发现以下错误:

scala> class EnhanceAny(x: AnyVal) { def squared = x * x }
<console>:7: error: value * is not a member of AnyVal
       class EnhanceAny(x: AnyVal) { def squared = x * x }

我很想知道,我如何能够将其应用于任何数字类别,或甚至适用于支持<代码>*的任何类别。

Answer 1

不可能找到任何类型的解决办法,采用* 方法,而不必为你想要处理的每一种类型写出一种碎块转换。从根本上说,你需要一种恢复性的结构性类型,而Schala则不支持这种结构,因为科索沃机械设备类型消失。详情见。

您可以与以下几类人相当接近:<条码>;Numeric 类型类别;。它将与多数阶层合作:

//define the type class
trait Multipliable[X] { def *(x: X): X}

//define an implicit from A <% Numeric[A] -> Multipliable[Numeric[A]]
implicit def Numeric2Mult[A](a: A)(implicit num: Numeric[A]): Multipliable[A] = new Multipliable[A]{def *(b: A) = num.times(a, b)}

//now define your Enhanced class using the type class
class EnhancedMultipliable[T <% Multipliable[T]](x: T){ def squared = x * x}

//lastly define the conversion to the enhanced class
implicit def Mult2EnhancedMult[T <% Multipliable[T]](x: T) = new EnhancedMultipliable[T](x)

3.squared
//Int = 9

3.1415F.squared
//Float = 9.869022

123456789L.squared
//Long = 15241578750190521

Answer 2

Go visit Scala: How to define “generic” function parameters?

Answer 3

Think about it. Multiplication have different properties on different objects. Integer multiplication, for example, is associative and commutative, scalar multiplication is not commutative. And multiplication on floating point numbers... But, of course, you can have what you want with trait or with "trait and implicit" construction which resembles a typeclass.

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