Switch statement: must default be the last case?

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

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.

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 are cases with a defined set of input values where there are 'exceptions' and normal cases. 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.

Another reason is that the order of the cases may change the compiled code behavior, and that matters for performances. Most compilers will generate compiled assembly code in the same order as the code appears in the switch. That makes the first case very different from the others: all cases except the first one will involve a jump and that will empty processor pipelines. You may understand it like branch predictor defaulting to running the first appearing case in the switch. If a case if much more common that the others then you have very good reasons to put it as the first case.

Reading comments it's the specific reason why the original poster asked that question after reading Intel compiler Branch Loop reorganisation about code optimisation.

Then it will become some arbitration between code readability and code performance. Probably better to put a comment to explain to future reader why a case appears first.