switch () {default: <something>; case ...

Consider the following:

Code:

---

#include <stdio.h>


int main(void)
{
        int i = 3;

        switch (i){
                default:
                        i = -1;
                        break;
                case 1:
                case 2:
                case 3:
                        i = 5;
                        break;
        }

        printf ("i = %i.\n", i);
        return 0;
}

---

Output is:
i = 5.

Is this a definition of the standard, or is this compiler specific (ie, gcc). This code example is one that I tried and it works as if the default was at the end, but I would think it would match the default FIRST and never test against 1,2,3.

The code that coRnfused me was from the Linux kernel in the file kernel/posix-cpu-timers.c.

Thanks in advance for your answers.

Andy

Quote:

---

Originally Posted by Kennedy

Is this a definition of the standard, or is this compiler specific (ie, gcc).

---

Well...

Quote:

---

Originally Posted by C99 Clause 6.8.4.2 Paragraph 5 (part)

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.

---

Hence the behaviour follows that described by the standard.

Hmm, so then everyone that I polled here believed it to be a syntax error, however, it worked, based upon the standard. So, this leads me to believe that near everyone is taught that default must come at the end, which is erroneous.

Yet again, I have found something I was taught to be fact as false.

Bummer man.

Thanks laserlight!

Andy

It's only stylistic convention that places it at the end.

For extra fun, mis-spell it as defualt and see how long it takes people to spot it ;)

> Hmm, so then everyone that I polled here believed it to be a syntax error
Better not show them Duff's device then, they'd all explode!