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如何将函数应用于元组？

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这应该是一个简单的 . 如何将函数应用于Scala中的元组？即：

scala> def f (i : Int, j : Int) = i + j
f: (Int,Int)Int

scala> val p = (3,4)
p: (Int, Int) = (3,4)

scala> f p
:6: error: missing arguments for method f in object $iw;
follow this method with `_' if you want to treat it as a partially applied function
       f p
       ^

scala> f _ p
:6: error: value p is not a member of (Int, Int) => Int
       f _ p
           ^

scala> (f _) p
:6: error: value p is not a member of (Int, Int) => Int
       (f _) p
             ^

scala> f(p)
:7: error: wrong number of arguments for method f: (Int,Int)Int
       f(p)
       ^

scala> grr!

提前谢谢了 .

scala

3 回答

  • 10

    在Scala 2.7中：

    scala> def f (i : Int, j : Int) = i + j
f: (Int,Int)Int

scala> val ff = f _
ff: (Int, Int) => Int = <function>

scala> val fft = Function.tupled(ff)
fft: ((Int, Int)) => Int = <function>

    在Scala 2.8中：

    scala> def f (i : Int, j : Int) = i + j
f: (i: Int,j: Int)Int

scala> val ff = f _
ff: (Int, Int) => Int = <function2>

scala> val fft = ff.tupled
fft: ((Int, Int)) => Int = <function1>
    回复于
  • 0

    跟进另一个答案，可以写（用2.11.4测试）：

    scala> def f (i: Int, j: Int) = i + j
f: (i: Int, j: Int)Int

scala> val ff = f _
ff: (Int, Int) => Int = <function2>

scala> val p = (3,4)
p: (Int, Int) = (3,4)

scala> ff.tupled(p)
res0: Int = 7

    见def tupled: ((T1, T2)) ⇒ R：

    创建此函数的tupled版本：它接受单个scala.Tuple2参数，而不是2个参数 .

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  • 57
    scala> def f (i: Int, j: Int) = i + j
f: (i: Int, j: Int)Int
scala> val p = (3,4)
p: (Int, Int) = (3,4)
scala> val ft = (f _).tupled
ft: ((Int, Int)) => Int = <function1>
scala> ft apply(p)
res0: Int = 7
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