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功能平等概念

手机用户2502921001 发布于 2023-02-12 18:38

Scala中是否存在功能平等的概念？

scala> def foo(x: Int) = x + 1
foo: (x: Int)Int

scala> def bar(x: Int) = x + 1
bar: (x: Int)Int

所以我不能打电话==给一个功能.

scala> foo == bar
:10: error: missing arguments for method foo;
follow this method with `_' if you want to treat it as a partially applied
 function
              foo == bar
              ^

除了验证所有x的foo(x)== (不知道如何做到这一点),我还能如何验证相等性？bar(x)

1 个回答

你不会真的期望==产生除false以外的任何东西,因为==意味着"具有相同的内存地址",除非它是专门为该类定义的.毕竟,
```
scala> class A
scala> (new A) == (new A)
res0: Boolean = false
```
如果Function1确实尝试提供实现==,则不清楚它将检查什么.所有可能的投入都有相同的结果同样的行为？相同的字节码？

如果你想要一个具有良好定义的相等操作的函数类,你必须自己编写它:
```
class Adder(private val toAdd: Int) extends (Int => Int) {
  override def apply(n: Int) = n + toAdd

  override def equals(other: Any) = other match {
    case o: Adder => this.toAdd == o.toAdd
    case _ => false
  }
}
```
然后你可以这样做:
```
scala> val foo = new Adder(1)
foo: Adder = <function1>

scala> val bar = new Adder(1)
bar: Adder = <function1>

scala> foo(2)
res4: Int = 3

scala> foo == bar
res5: Boolean = true
```
要获得更简单的解决方案,只需将您的类设为a case class,即可免费获得equals和hashCode实现:
```
case class Adder(private val toAdd: Int) extends (Int => Int) {
  override def apply(n: Int) = n + toAdd
}
```
2023-02-12 18:40 回答

放肆的微笑-扯痛了忧伤

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