比较匹配器在混合数字类型上失败 [英] Comparison matchers fail on mixed numeric types

问题描述

在 vanilla Scala 中，以下断言通过

In vanilla Scala the following assertions pass

assert(1D > 0F)
assert(1F > 0)
assert(1L > 0)
assert(1 > 0.toShort)
assert(1.toShort > 0.toChar)

然而 ScalaTest 中的相似匹配器失败

however similar matchers in ScalaTest fail

1D shouldBe > (0F)
1F shouldBe > (0)
1L shouldBe > (0)
1 shouldBe > (0.toShort)
1.toShort shouldBe > (0.toChar)

一种解决方法是让双方都成为相同的类型，例如

A workaround is to make both sides the same type, for example

1D shouldBe > (0D)

为什么它在 Scala 中有效，而在 Scalatest 中无效，或者 >

Why does it work in Scala, but not in Scalatest, or what is it about the signature of >

def >[T : Ordering] (right: T): ResultOfGreaterThanComparison[T]

这让它失败了吗?

推荐答案

Vanilla Scala 工作由于自动类型转换，即 0F 被强制转换为 0D 这是一个常见的做法多种语言.

Vanilla Scala works due to automatic type conversion i.e. 0F is cast to 0D which is a common practise in many languages.

更有趣的问题是为什么 shouldBe 不起作用.去除隐含物的糖分

More interesting question is why shouldBe does not work. De-sugaring the implicits yields

new AnyShouldWrapper[Double](leftSideValue = 1D,
                             pos = ???,
                             prettifier = ???)
  .shouldBe(new ResultOfGreaterThanComparison[Double](right = 0D))

new AnyShouldWrapper[Double](leftSideValue = 1D,
                             pos = ???,
                             prettifier = ???)
  .shouldBe(new ResultOfGreaterThanComparison[Float](right = 0F))

导致 shouldBe 的重载实现.前一种情况 此处 和后者这里.

which leads to overloaded implementations of shouldBe. The former case goes here and the latter here.

看了源码，好像唯一的原因1D shouldBe >(0F) 实际编译是为了支持数组比较，加上shouldBe关键字.

After looking at the source code, it seems that the only reason 1D shouldBe > (0F) actually compiles is to support array comparison with shouldBe keyword.

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