一个 Switch Java 问题:case 表达式必须是常量表达式
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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