我不能将元组作为方法参数传递:
scala> val c:Stream[(Int,Int,Int)]= Stream.iterate((1,1))((a:Int,b:Int,c:Int) => (b,c,a+b)) <console>:11: error: type mismatch; found : (Int,Int) => (Int,Int) required: ((Int,Int)) => (Int,Int)
谢谢.
解决方法
就像函数文字一样:
(x:Int) => x + 1
(x:Int,y: Int,z: Int) => x + y + z
您可以使用case语句巧妙地完成这项工作:
scala> val c: Stream[(Int,Int)] = Stream.iterate((1,1)){ case (a,b,c) => (b,a+b) } c: Stream[(Int,Int)] = Stream((1,1),?)
另一种方法是传递元组,但由于所有的_1访问器,这真的很难看:
scala> val c:Stream[(Int,Int)] = Stream.iterate((1,1))( t => (t._2,t._3,t._1 + t._2) ) c: Stream[(Int,?)