foreach loop in scala

In scala foreach loop if I have list

val a = List("a","b","c","d")

I can print them without a pattern matching like this

a.foreach(c => println(c))

But, if I have a tuple like this

val v = Vector((1,9), (2,8), (3,7), (4,6), (5,5))

why should I have to use

v.foreach{ case(i,j) => println(i, j) }

a pattern matching case

{ brackets

Please explain what happens when the two foreach loops are executed.

---

You don't have to, you choose to. The problem is that the current Scala compiler doesn't deconstruct tuples, you can do:

v.foreach(tup => println(tup._1, tup._2))

But, if you want to be able to refer to each element on it's own with a fresh variable name, you have to resort to a partial function with pattern matching which can deconstruct the tuple.

This is what the compiler does when you use case like that:

def main(args: Array[String]): Unit = {
  val v: List[(Int, Int)] = scala.collection.immutable.List.apply[(Int, Int)](scala.Tuple2.apply[Int, Int](1, 2), scala.Tuple2.apply[Int, Int](2, 3));
  v.foreach[Unit](((x0$1: (Int, Int)) => x0$1 match {
    case (_1: Int, _2: Int)(Int, Int)((i @ _), (j @ _)) => scala.Predef.println(scala.Tuple2.apply[Int, Int](i, j))
  }))
}

You see that it pattern matches on unnamed x0$1 and puts _1 and _2 inside i and j , respectively.

---

To answer #2: You can only use case in braces. A more complete answer about braces is located here.

---

根据http://alvinalexander.com/scala/iterating-scala-lists-foreach-for- compprehension：

val names = Vector("Bob", "Fred", "Joe", "Julia", "Kim")

for (name <- names)
    println(name)

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