Understanding switch behaviour

Question

Consider the following code

int x = 1
switch(x) {
    case 1:
        System.out.println("1");
    case 2:
        System.out.println("2");
    default:
        System.out.println("no value found");
 }

it prints

1
2
no value found

as expected because there is no break in each case statement
my doubt is that how execution goes to each case statement if there is no break in first one
because x is not equal 2 then also its execute case 2 block
but i understood default one because without break program continue its execution and execute the default statement

Joey · Accepted Answer · 2013-07-21T08:20:06.843

This is a remnant of the C programming language. In C switch is little more than a little syntactic sugar over a goto depending on the expression value and thus execution simply jumps to the appropriate case and continues from there. The labels in between are exactly that: labels. They are jump targets but don't affect control flow otherwise and from the compiler's point of view there is nothing in a label that merits a jump elsewhere (except there is INTERCAL's COMEFROM somewhere). So you have to put an explicit break after every case where you don't want to fall through to other options.

Java more or less inherited those semantics while eschewing some of the crazier C idioms.

C# on the other hand goes a bit further by disallowing fall-through entirely on non-empty case labels.

In my opinion though, this was a bit of a design error. Fall-through by default may be much easier to implement in the compiler (because you don't have to add a goto to the end of the switch and it aligns very nicely with how things actually work underneath), but in my experience there are vastly more programming errors by accidentally using fall-through than by accidentally not using it.

The most severe design error in my eyes is including the switch statement with all its weirdness in modern languages.

score 0 · Answer 2 · answered Jul 21 '13 at 08:12

0

For reasons that are inherited from C++, execution naturally proceeds until the end of switch or a break. There is no test against cases as they are encountered.

answered Jul 21 '13 at 08:12

tbodt

16,609
6
58
83

score 0 · Answer 3 · answered Jul 21 '13 at 08:12

switch is fall through. You need to add break before each next case unless you want that.

switch (x) {
    case 1:
        System.out.println("1");
        break;
    case 2:
        System.out.println("2");
        break;
    default:
        System.out.println("no value found");
}

score 0 · Answer 4 · edited Jun 20 '20 at 09:12

The switch statement is an abstraction of a branch table, and a branch table also has an implicit fall-through and requires an additional jump instruction to avoid it. Read this SO answer as well.

As per JLS 14.11:

The body of a switch statement is known as a switch block.

If one of the case constants is equal to the value of the expression, then we say that the case matches, and all statements after the matching case label in the switch block, if any, are executed in sequence.

If all these statements complete normally, or if there are no statements after the matching case label, then the entire switch statement completes normally.

score 0 · Answer 5 · answered Jul 21 '13 at 08:13

0

It's to do with how it's compiled and optimization.

It's why a switch is preferable to a bunch of chained if statements.

answered Jul 21 '13 at 08:13

j00lz

281
1
12

score 0 · Answer 6 · answered Jul 21 '13 at 08:13

From java doc:

The break statements are necessary because without them, statements in switch blocks fall through: All statements after the matching case label are executed in sequence, regardless of the expression of subsequent case labels, until a break statement is encountered

score 0 · Answer 7 · answered Jul 21 '13 at 08:16

The switch statement is not like an if-then-else statement. switch statement has multiple execution path.

The Documentation states:

Control flow continues with the first statement following the switch block. The break statements are necessary because without them, statements in switch blocks fall through: All statements after the matching case label are executed in sequence, regardless of the expression of subsequent case labels, until a break statement is encountered.

score 0 · Answer 8 · answered Jul 21 '13 at 08:18

It's correct behaviour. "default" doesn't mean all possible answers. It means all answers without these written in cases.

int x=1
switch(x)
    {
        default:
             System.out.println("no value found");
        case 1:
            System.out.println("1");
        case 2:
            System.out.println("2");
    }

will print

1
2

won't print

no value found
1
2

when using breaks; statements and default value there is called only one "branch" for each value

score 0 · Answer 9 · answered Jul 21 '13 at 08:18

Each break statement terminates the enclosing switch statement. All statements after the matching case label are executed in sequence, regardless of the expression of subsequent case labels, until a break statement is encountered.

You can use if-then-else just, the overhead of testing is increases..., which is avoided in switch-case, but break is necessary for proper execution...

Understanding switch behaviour

9 Answers9