scala - 为什么 scala 不能将 Long switch 语句编译为 tableswitch

Question

当我这样做的时候

val oldId :Long  = 123;

val i = 1
val newId = (oldId: @switch) match {
  case 1  => 1234
  case 2  => 5678
  case 3  => 1122
  case 4  => 3344
}

我收到编译器警告

[ant:scalac] mycode.scala:25: warning: could not emit switch for @switch annotated match
[ant:scalac]     val newId = (oldId: @switch) match {
[ant:scalac]                          ^
[ant:scalac] one warning found

但是如果我使用以下代码

val oldId :Int  = 123;

val i = 1
val newId = (oldId: @switch) match {
  case 1  => 1234
  case 2  => 5678
  case 3  => 1122
  case 4  => 3344
}

然后编译器不会给我警告。为什么不能使用 Long 并获得 tableswitch？

Answer 1

Why is it not possible to use a Long and get a tableswitch?

因为 JVM supports @tableswitch on int and types convertable to int :

The Java Virtual Machine's tableswitch and lookupswitch instructions operate only on int data. Because operations on byte, char, or short values are internally promoted to int, a switch whose expression evaluates to one of those types is compiled as though it evaluated to type int

因此，您会收到编译时警告，因为 long 不能隐式转换为 int。