C switch statement: has default to be the last case?

Question

Consider the following switch statement:

switch( value )
{
  case 1:
    return 1;
  default:
    value++;
    // fall-through
  case 2:
    return value * 2;
}

This code compiles, but is it valid (= defined behaviour) for C90/C99? I have never seen code where the default case is not the last case.

EDIT:
As Jon Cage and KillianDS write: this is really ugly and confusing code and I am well aware of it. I am just interested in the general syntax (is it defined?) and the expected output.
Thanks for your contributions!

Answer 1

yes, this is valid, and under some circumstances it is even useful. Generally, if you don't need it, don't do it.

Answer 2

The case statements and the default statement can occur in any order in the switch statement. The default clause is an optional clause that is matched if none of the constants in the case statements can be matched.

Good Example :-

switch(5) {
  case 1:
    echo "1";
    break;
  case 2:
  default:
    echo "2, default";
    break;
  case 3;
    echo "3";
    break;
}


Outputs '2,default'

very useful if you want your cases to be presented in a logical order in the code (as in, not saying case 1, case 3, case 2/default) and your cases are very long so you do not want to repeat the entire case code at the bottom for the default

Answer 3

It's valid, but rather nasty. I would suggest it's generally bad to allow fall-throughs as it can lead to some very messy spaghetti code.

It's almost certainly better to break these cases up into several switch statements or smaller functions.

[edit] @Tristopia: Your example:

Example from UCS-2 to UTF-8 conversion 

r is the destination array, 
wc is the input wchar_t  

switch(utf8_length) 
{ 
    /* Note: code falls through cases! */ 
    case 3: r[2] = 0x80 | (wc & 0x3f); wc >>= 6; wc |= 0x800; 
    case 2: r[1] = 0x80 | (wc & 0x3f); wc >>= 6; wc |= 0x0c0; 
    case 1: r[0] = wc;
}

would be clearer as to it's intention (I think) if it were written like this:

if( utf8_length >= 1 )
{
    r[0] = wc;

    if( utf8_length >= 2 )
    {
        r[1] = 0x80 | (wc & 0x3f); wc >>= 6; wc |= 0x0c0; 

        if( utf8_length == 3 )
        {
            r[2] = 0x80 | (wc & 0x3f); wc >>= 6; wc |= 0x800; 
        }
    }
}   

[edit2] @Tristopia: Your second example is probably the cleanest example of a good use for follow-through:

for(i=0; s[i]; i++)
{
    switch(s[i])
    {
    case '"': 
    case '\'': 
    case '\\': 
        d[dlen++] = '\\'; 
        /* fall through */ 
    default: 
        d[dlen++] = s[i]; 
    } 
}

..but personally I would split the comment recognition into it's own function:

bool isComment(char charInQuestion)
{   
    bool charIsComment = false;
    switch(charInQuestion)
    {
    case '"': 
    case '\'': 
    case '\\': 
        charIsComment = true; 
    default: 
        charIsComment = false; 
    } 
    return charIsComment;
}

for(i=0; s[i]; i++)
{
    if( isComment(s[i]) )
    {
        d[dlen++] = '\\'; 
    }
    d[dlen++] = s[i]; 
}

Answer 4

The "default" condition can be anyplace within the switch that a case clause can exist. It is not required to be the last clause. I have seen code that put the default as the first clause. The "case 2:" gets executed normally, even though the default clause is above it.

As a test, I put the sample code in a function, called test(int value){} and ran:

  printf("0=%d\n", test(0));
  printf("1=%d\n", test(1));
  printf("2=%d\n", test(2));
  printf("3=%d\n", test(3));
  printf("4=%d\n", test(4));

The output is:

0=2
1=1
2=4
3=8
4=10

Answer 5

There's no defined order in a switch statement. You may look at the cases as something like a named label, like a goto label. Contrary to what people seem to think here, in the case of value 2 the default label is not jumped to. To illustrate with a classical example, here is Duff's device, which is the poster child of the extremes of switch/case in C.

send(to, from, count)
register short *to, *from;
register count;
{
  register n=(count+7)/8;
  switch(count%8){
    case 0: do{ *to = *from++;
    case 7:     *to = *from++;
    case 6:     *to = *from++;
    case 5:     *to = *from++;
    case 4:     *to = *from++;
    case 3:     *to = *from++;
    case 2:     *to = *from++;
    case 1:     *to = *from++;
            }while(--n>0);
  }
}

Answer 6

It's valid and very useful in some cases.

Consider the following code:

switch(poll(fds, 1, 1000000)){
   default:
    // here goes the normal case : some events occured
   break;
   case 0:
    // here goes the timeout case
   break;
   case -1:
     // some error occurred, you have to check errno
}

The point is that the above code is more readable and efficient than cascaded if. You could put default at the end, but it is pointless as it will focus your attention on error cases instead of normal cases (which here is the default case).

Actually, it's not such a good example, in poll you know how many events may occur at most. My real point is that there is cases with a defined set of input values where there is 'exceptions' and normal case. If it's better to put exceptions or normal cases at front is a matter of choice.

In software field I think of another very usual case : recursions with some terminal values. If you can express it using a switch, default will be the usual value that contains recursive call and distinguished elements (individual cases) the terminal values. There is usually no need to focus on terminal values.

Answer 7

The C99 standard is not explicit about this, but taking all facts together, it is perfectly valid.

A case and default label are equivalent to a goto label. See 6.8.1 Labeled statements. Especially interesting is 6.8.1.4, which enables the already mentioned Duff's Device:

Any statement may be preceded by a prefix that declares an identifier as a label name. Labels in themselves do not alter the flow of control, which continues unimpeded across them.

Edit: The code within a switch is nothing special; it is a normal block of code as in an if-statement, with additional jump labels. This explains the fall-through behaviour and why break is necessary.

6.8.4.2.7 even gives an example:

switch (expr) 
{ 
    int i = 4; 
    f(i); 
case 0: 
    i=17; 
    /*falls through into default code */ 
default: 
    printf("%d\n", i); 
} 

In the artificial program fragment the object whose identifier is i exists with automatic storage duration (within the block) but is never initialized, and thus if the controlling expression has a nonzero value, the call to the printf function will access an indeterminate value. Similarly, the call to the function f cannot be reached.

The case constants must be unique within a switch statement:

6.8.4.2.3 The expression of each case label shall be an integer constant expression and no two of the case constant expressions in the same switch statement shall have the same value after conversion. There may be at most one default label in a switch statement.

All cases are evaluated, then it jumps to the default label, if given:

6.8.4.2.5 The integer promotions are performed on the controlling expression. The constant expression in each case label is converted to the promoted type of the controlling expression. If a converted value matches that of the promoted controlling expression, control jumps to the statement following the matched case label. Otherwise, if there is a default label, control jumps to the labeled statement. If no converted case constant expression matches and there is no default label, no part of the switch body is executed.