Java switch : variable declaration and scope

问题:

How does the Java compiler handle the following switch block ? What is the scope of the 'b' variable ?

Note that the 'b' variable is declared only in the first branch of the switch statement. Attempting to declare it in the second branch as well results in a "duplicate local variable" compilation error.

    int a = 3;
    switch( a ) {
    case 0:
        int b = 1;
        System.out.println("case 0: b = " + b);
        break;
    case 1:
        // the following line does not compile: b may not have been initialized
        // System.out.println("case 1 before: b = " + b);
        b = 2;
        System.out.println("case 1 after: b = " + b);
        break;
    default:
        b = 7;
        System.out.println("default: b = " + b);
    }

Note: the above code compiles with a java 1.6 compiler.

回答1:

The scope is, just as usual, delimited by { and }.

回答2:

The scope of b is the block. You have only one block which includes all cases. That's why you get a compile error when you redeclare b in your second case.

You could wrap each case in an own block like

case 0:
   {
     int b = 1;
     ...
   }
case 1:
   {
     int b = 2;
     ...
   }

but I think FindBugs or CheckStyle would complain about that.

回答3:

The scope of b is the switch block - between the declaration and the delimiter } -

int a = 3;

switch( a ) {
     case 0:
          int b = 1; //scope starts
          System.out.println("case 0: b = " + b);
          break;
     case 1:
          // the following line does not compile: b may not have been initialized
          // System.out.println("case 1 before: b = " + b);
          b = 2;
          System.out.println("case 1 after: b = " + b);
          break;
     default:
          b = 7;
          System.out.println("default: b = " + b);
}//scope ends

However, you need to know that if you declare the int b inside the case 1:, you will NOT have access to the variable b inside the case 0:

To answer the question you ask in the java comments you can check this simpler example:

int b;
if(true){
    b++; //The local variable b hast not been initialized
}

Hope it helps.

回答4:

You can define the scope using {} around your case.

int a = 3;
switch( a ) {
case 0: {
    int b = 1;
    System.out.println("case 0: b = " + b);
    break;
}
case 1: {
    // the following line does not compile: b may not have been initialized
    // System.out.println("case 1 before: b = " + b);
    int b = 2;
    System.out.println("case 1 after: b = " + b);
    break;
}
default: {
    int b = 7;
    System.out.println("default: b = " + b);
}
}

回答5:

in your code if a is not equal to 0 b will never be initialized. you should define b before the switch statement.

回答6:

Your case blocks do not have any local scope. It's not a series of if...else if...else blocks, java implements it as a series of GOTOs.

回答7:

The Scope of variables defined in a switch() statement would be the same as in a normal block wich is surround by { and }.

Therefore every variable defined in a switch() statement is visible for the whole block, once it is defined.