switch statement

[Angela lewis]

 class Test
{
public static void main(String[] args) {
char x = '1';
switch (x) {
case 1: {
System.out.println("x is equal to 1");
break;
}
case 2:
System.out.println("x is equal to 2");
break;
case 3:
System.out.println("x is equal to 3");
break;
default:
System.out.println("default");
break;
}
}
}

Why does this code produce "default" and not 1 ?

[Geoffrey Vlassaks]

That's because the unicode value of '1' will be taken to compare in the switch statement. This unicode value differs from 1, 2 and 3 so the default will be taken...

Greetz

[Wendal Park]

Geoffrey, I believe your explanation isn't quite correct.

The integer value of a char primitive is in fact the ASCII value of the character it represents. As it shows in the following code:

char x = 'a';
short s = (short)x;
System.out.println(s);
// This will print 97, which is the ASCII value of 'a'

So even for char x = '1', the value x holds is the ASCII value of '1', which is 49.
[ May 25, 2004: Message edited by: Wendal Park ]

[Barry Gaunt]

Wendal, the char does not have to be a ASCII character like 'a' or '1', it could be some value like \u0546 (whatever that is) which is way out of the ASCII range.

public class TestMe {
public static void main(String[] args) {
char a = '\u0546';
switch (a) {
case 1280 + 64 + 6: {
System.out.println("found character: " + a);
break;
}
default:
System.out.println("unknown");
}
}
}

[ May 25, 2004: Message edited by: Barry Gaunt ]

[Wendal Park]

Oh right, thanks for pointing out that.
I was just focusing on the single char (e.g. 'a') case which made me think of ASCII from the very beginning.