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.
24 Answers
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.
According to :
val names = Vector("Bob", "Fred", "Joe", "Julia", "Kim")
for (name <- names) println(name)To answer #2: You can only use case in braces. A more complete answer about braces is located here.
Vector is working a bit differently, you using function literals using case...
In Scala, we using brackets{} which accept case...
{ case pattern1 => "xxx" case pattern2 => "yyy"
}So, in this case, we using it with foreach loop...
Print all values using the below pattern then:
val nums = Vector((1,9), (2,8), (3,7), (4,6), (5,5))
nums.foreach { case(key, value) => println(s"key: $key, value: $value")
}Also you can check other loops like for loop as well if you think this is not something which you are comfortable with...
