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# JDK6: about switch

Posted on 2010-11-27
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hi all,

the following codes give the output

9 10 10 d 13

but the variable s has actually value 9, 10 and 11.

can anyone explain how exactly here works?

Codes:

public class Ebb {

static int x = 7;
public static void main(String[] args) {
// TODO code application logic here
String s = "";
for (int y = 0; y <3; y++) {
x++;
switch(x) {
case 8: s += "8";
case 9: s += "9";
case 10: { s += 10; break;}
default: s += "d";
case 13: s+= "13";
}
}
System.out.println(s);
}

static {
x++;
}
}

thanks,

wantime
0
Question by:wantime
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Expert Comment

ID: 34221775
In a switch clause, every case is performed after the match until break in encountered.
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Accepted Solution

aaver earned 2000 total points
ID: 34221807
In more detail:
In the loop x has the values 9, 10, and 11.
x = 9 -> the match is at case: 9 -> s += "9" // s = "9"
since no break is encountered the case: 10 is also performed -> s += 10 // s = "9" + 10 = "910"
now a break is encountered and the loop continues with the next value

x = 10 -> the match is at case: 10 -> s += 10 // s = "910" + 10 = "91010"
a break is encountered and the loop continues with the next value

x = 11 -> there is no match -> the match is at default:-> s += "d" // s = "91010" + "d" = "91010d"
since no break is encountered the case: 13 is also performed -> s += "13" // s = "91010d" + "13" = "91010d13"
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