Java SE 7 switch语句，默认情况何时执行？

Question

I'm preparing OCA Java SE 7 example. A question about the switch statement: here is the code:

int a = 5;
switch(a){
    default: //print 0
    case 1: //print 1
    case 5: //print 5
    case 8: //print 8
}

what's the output of the code? Here is something I don't understand: the default will usually be the last condition. What if default is the first condition? will the default case been matched first?

one step further: if I change the code, make i = 10, which certainly won't match the rest 3 condition, but will match the default condition. Given that no break; in my code, will the program output : 0 1 5 8 ??

update I don't think I state my question clear... As been suggested, I tried it with code and I do understand the rules about fall through.

And if i = 10, the result will be: 0 1 5 8.

what confused me is that: if i = 10

and when we enter the switch statement, which line executed first?

default: //print 0

or

case 1: //print 1

update update thank you guys... I think I get it...

the default case will only be triggered when there is no matches.

Which means in this case, the execution sequence should be:

case 1:
case 5:
case 8:
//and no found yet..trigger the default case
// given that my default case is at the top and no break;
//it fall through and print: 0 1 5 8

FYI: I do know how to run the code. And I didn't expect such a huge volume of criticisms saying that "do it yourself", "don't just read and no practice"..

anyway, thanks.

Answer 1

You should try it and see. But one thing I'll point out is that there are no break statements. This means it will fall through from where ever it ends up.

So what will happen is it'll evaluate the switch and jump to the appropriate case. From there it will fall through.

Answer 2

switch is just a specific goto pattern deemed safe for general use.

Gives something like this:

if (i == 1) goto lbl1;
if (i == 5) goto lbl2;
if (i == 8) goto lbl3;
lbl4: print 0;
Lbl1: print 1;
Lbl2: print 5;
Lbl3: print 8;