一个 Switch Java 问题:case 表达式必须是常量表达式 [英] A Switch Java problem : case expressions must be constant expressions

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://docs.oracle.com/javase/specs/jls/se7/html/jls-14.html#jls-14.11 了解 switch 的正式定义.

See: http://docs.oracle.com/javase/specs/jls/se7/html/jls-14.html#jls-14.11 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.

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