问题
When I run something like the following from the command line, what really happens?
> scala hello.scala
Is there a hello.class generated, executed, and then discarded? Or does Scala behave somehow like an interpreter in this case? I am just thinking that, of course, I cannot do the same for Java:
> java hello.java
回答1:
Yes, there is a hello.class generated. The compiler will wrap your code inside a Main object, compile it then execute Main.main, given hello.scala of
println(args.mkString)
println(argv.mkString)
If you run with the -Xprint:parser option: scala -Xprint:parser hello.scala foo bar you'll see how the code gets rewritten:
package <empty> {
object Main extends scala.ScalaObject {
def <init>() = {
super.<init>();
()
};
def main(argv: Array[String]): scala.Unit = {
val args = argv;
{
final class $anon extends scala.AnyRef {
def <init>() = {
super.<init>();
()
};
println(args.mkString);
println(argv.mkString)
};
new $anon()
}
}
}
}
This code is then compiled (I believe to a memory filesystem - but I'm not sure) and executed. Looking at ScriptRunner, I see that a temporary directory is created under the default temp folder. For instance looking at my system, I see a bunch of %TEMP%/scalascript* folders.
Note that even in the interpreter, the code is not interpreted. See Scala: Is there a default class if no class is defined? for more info (it's really being rewritten, compiled and evaluated).
来源：https://stackoverflow.com/questions/7655165/what-really-happens-behind-the-scala-runtime-repl-when-running-a-scala-progra