Question

我有以下代码：

def myfunct(n: Int, steps: Int) = n match {
  case 1 =>  steps
  case (x) => if (x % 2 == 0) ...

无论如何将偶数匹配逻辑移到案例中？我需要一个案例类吗？

如：

def myfunct(n: Int, steps: Int) = n match {
  case 1 =>  steps
  case (even number??) => ...

Answer 1

是的，它被称为警卫：

object One {
  def unapply(n: Int) = if (n == 1) Some(1) else None
}

object Even {
 def unapply(n: Int) = if (n % 2 == 0) Some(n) else None
}

def myfunct (n: Int, steps: Int) = n match {
 case One(x) =>  steps
 case Even(x) => doSomething
 case x => doSomething
}

Answer 2

您也可以使用提取器：

object Even {
  def unapply(x: Int) = if (x % 2 == 0) Some(x) else None
}

object Odd {
  def unapply(x: Int) = if (x % 2 == 1) Some(x) else None
}

List(1,2,3,4).foreach { 
  case Even(x) => println(s"$x: even")
  case Odd(x) => println(s"$x: odd")
}

Answer 3

使用Patten Matching guards

// On Quantity Change:
var qtys = document.getElementsByName("qty[]");
var i;
for (i = 0; i < qtys.length; ++i) {
    qtys[i].addEventListener("change", totalIt, false);
    qtys[i].addEventListener("input", totalIt, false);
}

其他方式是定义def myfunct (n: Int, steps: Int) = n match { case 1 => steps case x if x % 2 == 0 => doSomething case x => doSomething }和Even提取器

Odd

Answer 4

n及其mod上的模式匹配，tupled，

def f(n: Int, steps: Int) = (n, n % 2) match {
  case (1, _) => steps
  case (_, 0) => steps + 2
  case _      => steps + 1
}

如何在scala的情况下模式匹配偶数？

4 个答案: