Scala奇怪的隐式拳击转换错误

问题描述：

有人可以告诉我以下原因为何不起作用吗?

Can someone tell me why the following does not work?

object TestObject {
    def map(f: (Double, Double) => Double, x2: Array[Double]) = {
        val y = x2.zip( x2 )
        val z = y.map(f)
        z
    }
}

产生此错误:

type mismatch; found : (Double, Double) => Double required: ((Double, Double)) => ?

答

在此代码段中，f是一个带有两个Double参数并返回Double的函数. 您正在尝试通过传递类型为Tuple2[Double,Double]的单个参数来调用f. 您可以通过首先更改f的类型来解决此问题:

In this snippet, f is a function taking two Double parameters and returning a Double. You are attempting to call f by passing a single argument of type Tuple2[Double,Double]. You can fix this by changing the type of f in the first place:

object TestObject {
    def map(f: ((Double, Double)) => Double, x2: Array[Double]) = {
        val y = x2.zip( x2 )
        val z = y.map(f)
        z
    }
}

您也可以将其声明为f: Tuple2[Double, Double] => Double以便更清楚(这是完全等效的).

You could also declare it as f: Tuple2[Double, Double] => Double to be clearer (this is entirely equivalent).

相反，您可以这样更改通话:

Conversely, you could change your call like this:

object TestObject {
    def map(f: (Double, Double) => Double, x2: Array[Double]) = {
        val y = x2.zip( x2 )
        val z = y.map(f.tupled)
        z
    }
}

tupled自动将您的(Double, Double) => Double函数转换为Tuple2[Double, Double] => Double函数. 但是请记住，每次调用TestObject.map

tupled automagically transforms your (Double, Double) => Double function into a Tuple2[Double, Double] => Double function. Keep in mind however that the conversion will be done on each call to TestObject.map

Scala奇怪的隐式拳击转换错误

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