我如何打破一个循环?
var largest=0
for(i<-999 to 1 by -1) {
for (j<-i to 1 by -1) {
val product=i*j
if (largest>product)
// I want to break out here
else
if(product.toString.equals(product.toString.reverse))
largest=largest max product
}
}
如何将嵌套for循环转换为尾递归?
来自FOSDEM 2009 Scala Talk http://www.slideshare.net/Odersky/fosdem-2009-1013261 第22页:
中断并继续 Scala没有它们。为什么? 它们有点迫不得已;最好使用许多较小的函数 发布如何与闭包交互。 他们是不需要的!
怎么解释呢?
当前回答
我是Scala的新手，但是这样可以避免抛出异常和重复方法:
object awhile {
def apply(condition: () => Boolean, action: () => breakwhen): Unit = {
while (condition()) {
action() match {
case breakwhen(true) => return ;
case _ => { };
}
}
}
case class breakwhen(break:Boolean);
像这样使用它:
var i = 0
awhile(() => i < 20, () => {
i = i + 1
breakwhen(i == 5)
});
println(i)
如果你不想断:
awhile(() => i < 20, () => {
i = i + 1
breakwhen(false)
});
其他回答
// import following package
import scala.util.control._
// create a Breaks object as follows
val loop = new Breaks;
// Keep the loop inside breakable as follows
loop.breakable{
// Loop will go here
for(...){
....
// Break will go here
loop.break;
}
}
使用Break模块 http://www.tutorialspoint.com/scala/scala_break_statement.htm
第三方易碎包是一种可能的替代方案
https://github.com/erikerlandson/breakable
示例代码:
scala> import com.manyangled.breakable._
import com.manyangled.breakable._
scala> val bkb2 = for {
| (x, xLab) <- Stream.from(0).breakable // create breakable sequence with a method
| (y, yLab) <- breakable(Stream.from(0)) // create with a function
| if (x % 2 == 1) continue(xLab) // continue to next in outer "x" loop
| if (y % 2 == 0) continue(yLab) // continue to next in inner "y" loop
| if (x > 10) break(xLab) // break the outer "x" loop
| if (y > x) break(yLab) // break the inner "y" loop
| } yield (x, y)
bkb2: com.manyangled.breakable.Breakable[(Int, Int)] = com.manyangled.breakable.Breakable@34dc53d2
scala> bkb2.toVector
res0: Vector[(Int, Int)] = Vector((2,1), (4,1), (4,3), (6,1), (6,3), (6,5), (8,1), (8,3), (8,5), (8,7), (10,1), (10,3), (10,5), (10,7), (10,9))
再加上雷克斯·科尔的另一种回答:
(1c)你也可以在你的循环中使用守卫: Var和= 0 For (i <- 0到1000;如果sum<1000) sum += I
巧妙地使用find方法进行收集将为您提供帮助。
var largest = 0
lazy val ij =
for (i <- 999 to 1 by -1; j <- i to 1 by -1) yield (i, j)
val largest_ij = ij.find { case(i,j) =>
val product = i * j
if (product.toString == product.toString.reverse)
largest = largest max product
largest > product
}
println(largest_ij.get)
println(largest)
在scala中我们能做的就是
scala> import util.control.Breaks._
scala> object TestBreak {
def main(args : Array[String]) {
breakable {
for (i <- 1 to 10) {
println(i)
if (i == 5)
break;
} } } }
输出:
scala> TestBreak.main(Array())
1
2
3
4
5
