关于scala.math.Integral的问题 [英] Question about scala.math.Integral

问题描述

scala的方法mkNumericOps和mkOrderingOps .math.Integral 的用途以及如何使用它们?

What do methods mkNumericOps andmkOrderingOps of scala.math.Integral do and how can we use them?

我知道可以将函数和对象方法声明为implicit并用于隐式转换.但是我不明白为什么将 traits方法声明为implicit.

I understand that functions and object methods can be declared implicit and used for implicit conversion. However I do not understand why traits methods are declared implicit.

顺便说一句，类方法也可以声明为implicit吗?

BTW, can class methods be declared implicit too?

推荐答案

首先，让我们看看它们的声明:

First, let's see their declaration:

implicit def mkNumericOps (lhs: T): IntegralOps
implicit def mkOrderingOps (lhs: T): Ops

它们是隐式的事实意味着它们的目标是提供一些自动值或转换.请注意，它们都从T转换为其他类型，其中T是特征的类型参数:Integral[T].

The fact that they are implicit means their goal is to provide some automatic value or conversion. Note that they both convert from T to some other type, where T is the type parameter of the trait: Integral[T].

因此，如果您有Integral[Int]，则mkNumericOps将为您提供从Int到IntegralOps的自动转换.这意味着您将能够从Int上的IntegralOps或Ops调用方法(或任何Integral的类型).

So, if you have Integral[Int], then mkNumericOps will give you an automatic conversion from Int to IntegralOps. That means you'll be able to call methods from IntegralOps or Ops on an Int (or whatever it is the type of your Integral).

现在，让我们看看这些是什么方法:

Now, let's see what methods are these:

def % (rhs: T): T
def * (rhs: T): T
def + (rhs: T): T
def - (rhs: T): T
def / (rhs: T): T
def /% (rhs: T): (T, T)
def abs (): T
def signum (): Int
def toDouble (): Double
def toFloat (): Float
def toInt (): Int
def toLong (): Long
def unary_- (): T

这些来自IntegralOps，扩展了Ops.关于它们的一个有趣的事情是，其中许多已经在Int上定义了！那么，为什么以及为什么会使用它们呢?这是一个示例:

These are from IntegralOps, which extends Ops. An interesting thing about them is that many of them are already defined on Int! So, how and why one would use them? Here's an example:

def sum[T](list: List[T])(implicit integral: Integral[T]): T = {
    import integral._   // get the implicits in question into scope
    list.foldLeft(integral.zero)(_ + _)
}

因此，给定任何类型T隐式可用的Integral[T]，您可以将该类型的列表传递给sum.

So, given any type T for which there's an Integral[T] implicitly available, you can pass a list of that type to sum.

如果另一方面，如果我将我的方法专用于类型Int，则可以在没有Integral的情况下编写该方法.另一方面，我无法编写对Int，Long和BigInt都适用的内容，因为它们没有共享定义方法+的共同祖先(更不用说"0"了)

If, on the other hand, I made my method specific for the type Int, I could write it without Integral. On the other hand, I can't write something that will work for both Int and Long and BigInt, because they do not share a common ancestor defining the method + (much less a `zero´).

上面的foldLeft有效翻译为:

list.foldLeft(integral.zero)((x, y) => mkNumericOps(x).+(y))

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