val list=List(1,2,3,4) list.reduce((x:Int,y:Int)=>x+y)--->list.reduceLeft((x:Int,y:Int)=>x+y) var first = true var acc:Int = 0 op=(x:Int,y:Int)=>x+y
for循环第一次循环：acc=1 first = false第二次循环：acc=op(1,2)=1+2=3第三次循环：acc=op(3,3)=3+3=6第四次循环：acc=op(6,4)=6+4=10====================================================================== val list=List(1,2,3,4) list.reduceRight((x:Int,y:Int)=>x-y) op(head, tail.reduceRight(op))---》op(1,List(2,3,4).reduceRight(op)---->op(2,List(3,4).reduceRight(op)-->op(3,List(4).reduceRight(op)=4)=3-4=-1)=2-(-1)=3)=1-3=-2 ======================================================================== val list=List(1,2,3,4) list.foldRight(0)(_-_) --->reverse.foldLeft(z)((right, left) => op(left, right)) List(4,3,2,1).foldLeft(z)((right, left) => op(left, right)) var acc = 0 var these = List(4,3,2,1) while (!these.isEmpty) { acc = op(acc, these.head) these = these.tail }while循环第一次while： acc=op(0,4)=>op(4,0)=4-0=4 these=List(3,2,1)第二次while： acc=op(4,3)=>op(3,4)=3-4=-1 these=List(2,1)第三次while： acc=op(-1,2)=>op(2,-1)=2-(-1)=3 these=List(1)第四次while： acc=op(3,1)=>op(1,3)=1-3=-2 these=List()
=========================================================================val array=Array(1,2,3,4)------>其本质是调用了Array这个object的apply方法
def apply(x: Int, xs: Int*): Array[Int] = { //构建了一个跟目标数组长度一致的空的数组 val array = new Array[Int](xs.length + 1) //把参数中的第一个元素赋值给空的数组的0下标 array(0) = x var i = 1 for (x <- xs.iterator) { array(i) = x; i += 1 } // array(1)=2 i=2 // array(2)=3 i=3 // array(3)=4 i=4 array--->Array(1,2,3,4) } import scala.collection.mutable._ Array(1,2,3,4).map(_+_) import scala.collection.mutable.A import scala.collection.mutable.B import scala.collection.mutable.C import scala.collection.mutable.D
转载于:https://www.cnblogs.com/mediocreWorld/p/11361164.html
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