switch : break without a case

Question

I have the following switch with a break without a case :

char c = 'a';
switch(c) {
    break;
    case 'a' : cout << 'a' << endl;
    break;
    case 'b' : cout << 'b' << endl;
    break;
    default : break;
}

Why does this code snippet print a? Shoudn't the switch break after encountering the first break statement only?

`switch` and `break` doesn't work like that (as you noticed). Statements not in `case` labels are simply disregarded. — Some programmer dude, Jan 27 '20 at 11:47
No, why? In switch you enter the code once appropriate case is encountered. Else default is trigerred. Why would that case-less break be even considered? It should be a warning when case-less code is in switch, tho. — ALX23z, Jan 27 '20 at 11:47
fwiw, i was expecting that this can be answered by reading https://en.cppreference.com/w/cpp/language/break, though after reading that one can still believe that your code does what you expect... — 463035818_is_not_an_ai, Jan 27 '20 at 12:06

score 4 · Accepted Answer · answered Jan 27 '20 at 11:49

4

The first break is ignored.

At switch(c), the runtime jumps to case 'a' as c=='a'. Everything before that is ignored.

This is why case 'b' works without printing 'a', and this is why variable definitions are frowned upon in switch blocks:

int n=0;
switch (n)
{
    int k=n;
case 0:
    return k; // UB
}

answered Jan 27 '20 at 11:49

YSC

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It seems to be ill-formed, not UB. I get `error: cannot jump from switch statement to this case label` `note: jump bypasses variable initialization`. – HolyBlackCat Jan 27 '20 at 11:52
@HolyBlackCat I _think_ this is UB but all major compilers are nice enough to raise an error; I see no mention of variables having different lifetime in `switch` blocks: http://eel.is/c++draft/stmt.switch – YSC Jan 27 '20 at 11:55
Also, nothing about `switch` here: http://eel.is/c++draft/basic.lookup – YSC Jan 27 '20 at 11:56
Hmm. I asked [a question](https://stackoverflow.com/questions/59931168/is-jumping-over-a-variable-initialization-ill-formed-or-causes-ub) about it. – HolyBlackCat Jan 27 '20 at 12:26

score 1 · Answer 2 · edited Jun 20 '20 at 09:12

1

In C, a case inside a switch is (analogous to) a goto label. The head of the switch is analogous to the goto itself.

Consequences

break must in general be used because the "switch-in" is goto-based ie not structured or single point of entry&exit
Any statement before the first case is never reached
Famous usage Duffs Device

edited Jun 20 '20 at 09:12

Community

1
1

answered Jan 27 '20 at 11:57

Rusi

1,054
10
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score 1 · Answer 3 · answered Jan 27 '20 at 12:05

A switch is a compact and readable form of what could be a series of ugly ifs and gotos otherwise:

char c = 'a';
if(c=='a')goto _a;         // test for "case 'a':"
if(c=='b')goto _b;         // test for "case 'b':"
goto _default;             // unconditional jump to "default:"
goto _break;               // <-- program never goes to this line, the one you are asking about
_a: cout << 'a' << endl;
goto _break;
_b: cout << 'b' << endl;
goto _break;
_default:
goto _break;
_break:

Test: https://ideone.com/OgsQuT

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

From cppreference:

If condition evaluates to the value that is equal to the value of one of constant_expressions, then control is transferred to the statement that is labeled with that constant_expression.

If condition evaluates to the value that doesn't match any of the case: labels, and the default: label is present, control is transferred to the statement labeled with the default: label.

The switch works as if there was a goto from the condition to the cases, hence your break is never executed. More generally anything you put there is dead code:

switch(1) {
    foo();                // dead code !
    case 1 : cout << '1'; // prints "1"
         break;       // and exits the switch
    case 2 : cout << '2';
         break;
}

switch : break without a case

4 Answers4

Consequences