问题描述
我的 switch/case 语句有问题.错误说:案例表达式必须是常量表达式".我理解错误,我可以使用 If 解决它,但有人可以告诉我为什么 case 表达式必须在 switch/case 中保持不变.我的错误代码示例:
I having a problem in my switch/case statement. The error says : "Case expressions must be constant expressions". I understand the error and I can resolve it using If but can someone tells me why the case expression must be constant in a switch/case. A code example of my error :
public boolean onOptionsItemSelected(MenuItem item) { int idDirectory = ((MenuItem) findViewById(R.id.createDirectory)).getItemId(); int idSuppression = ((MenuItem) findViewById(R.id.recycleTrash)).getItemId(); int idSeeTrash = ((MenuItem) findViewById(R.id.seeTrash)).getItemId(); switch (item.getItemId()) { case idDirectory: createDirectory(currentDirectory); break; case idSuppression: recycleTrash(); break; case idSeeTrash: seeTrash(); break; } return super.onOptionsItemSelected(item); }
谢谢你的解释!!
推荐答案
所以可以在编译阶段求值(静态检查)
So it can be evaluated during the compilation phase ( statically check )
参见:http://www.51sjk.com/Upload/Articles/1/0/337/337618_20221130103136835.11 了解 switch 的正式定义.
See: http://www.51sjk.com/Upload/Articles/1/0/337/337618_20221130103136835.11 for a formal definition of the switch.
此外,它还可以帮助您更好地理解 switch 是如何转换为字节码的:
Additionally it may help you to understand better how that switch is transformed into bytecode:
class Switch { void x(int n ) { switch( n ) { case 1: System.out.println("one"); break; case 9: System.out.println("nine"); break; default: System.out.println("nothing"); break; } } }
编译后:
C:>javap -c Switch Compiled from "Switch.java" class Switch extends java.lang.Object{ Switch(); Code: 0: aload_0 1: invokespecial #1; //Method java/lang/Object."<init>":()V 4: return void x(int); Code: 0: iload_1 1: lookupswitch{ //2 1: 28; 9: 39; default: 50 } 28: getstatic #2; //Field java/lang/System.out:Ljava/io/PrintStream; 31: ldc #3; //String one 33: invokevirtual #4; //Method java/io/PrintStream.println:(Ljava/lang/String;)V 36: goto 58 39: getstatic #2; //Field java/lang/System.out:Ljava/io/PrintStream; 42: ldc #5; //String nine 44: invokevirtual #4; //Method java/io/PrintStream.println:(Ljava/lang/String;)V 47: goto 58 50: getstatic #2; //Field java/lang/System.out:Ljava/io/PrintStream; 53: ldc #6; //String nothing 55: invokevirtual #4; //Method java/io/PrintStream.println:(Ljava/lang/String;)V 58: return }
看到标记为1的那一行:
1: lookupswitch{ //2 1: 28; 9: 39; default: 50 }
它评估值并转到其他行.例如,如果值为 9,它将跳转到指令 39:
It evaluates the value and goes to some other line. For instance if value is 9 it will jump to instruction 39:
39: getstatic #2; //Field java/lang/System.out:Ljava/io/PrintStream; 42: ldc #5; //String nine 44: invokevirtual #4; //Method java/io/PrintStream.println:(Ljava/lang/String;)V 47: goto 58
然后跳转到指令 58:
Which in turn jumps to instruction 58 :
58: return
如果动态评估,这一切都是不可能的.这就是为什么.
All this wouldn't be possible if it was evaluated dynamically. That's why.