Scala : it is impossible to put a tuple as a function's argument

Question

我不能将元组作为方法参数传递:

scala> val c:Stream[(Int,Int,Int)]= Stream.iterate((1, 0, 1))((a:Int,b:Int,c:Int) => (b,c,a+b))
<console>:11: error: type mismatch;
 found   : (Int, Int, Int) => (Int, Int, Int)
 required: ((Int, Int, Int)) => (Int, Int, Int)

谢谢。

Answer 1

就像函数文字一样:

(x:Int) => x + 1

是一个参数的函数，如下

(x:Int, y: Int, z: Int) => x + y + z

是三个参数的函数，而不是 3tuple 的一个参数

您可以使用 case 巧妙地完成这项工作。陈述:

scala> val c: Stream[(Int,Int,Int)] = 
       Stream.iterate((1, 0, 1)){ case (a, b, c) => (b, c, a+b) }

c: Stream[(Int, Int, Int)] = Stream((1,0,1), ?)

另一种方法是传递元组，但由于所有 _1，这真的很难看。访问器(accessor):

scala> val c:Stream[(Int,Int,Int)] = 
         Stream.iterate((1, 0, 1))( t => (t._2, t._3, t._1 + t._2) )

c: Stream[(Int, Int, Int)] = Stream((1,0,1), ?)