Scala: Is there a default class if no class is defined?
According to this, Scala methods belong to a class. However, if I define a method in REPL or in a script that I then execute using scala, what class does the method belong to ?
开发者_StackOverflow社区scala> def hoho(str:String) = {println("hoho " + str)}
hoho: (str: String)Unit
scala> hoho("rahul")
hoho rahul
In this example, what class does the method belong to ?
The REPL wraps all your statements (actually rewrites your statements) in objects automagically. You can see it in action if you print the intermediate code by using the -Xprint:typer
option:
scala> def hoho(str:String) = {println("hoho " + str)}
[[syntax trees at end of typer]]// Scala source: <console>
package $line1 {
final object $read extends java.lang.Object with ScalaObject {
def this(): object $line1.$read = {
$read.super.this();
()
};
final object $iw extends java.lang.Object with ScalaObject {
def this(): object $line1.$read.$iw = {
$iw.super.this();
()
};
final object $iw extends java.lang.Object with ScalaObject {
def this(): object $line1.$read.$iw.$iw = {
$iw.super.this();
()
};
def hoho(str: String): Unit = scala.this.Predef.println("hoho ".+(str))
}
}
}
}
So your method hoho
is really $line1.$read.$iw.$iw.hoho
. Then when you use hoho("foo")
later on, it'll rewrite to add the package and outer objects.
Additional notes: for scripts, -Xprint:typer
(-Xprint:parser
) reveals that the code is wrapped inside a code block in the main(args:Array[String])
of an object Main
. You have access to the arguments as args
or argv
.
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