Faulty default!

[Anjali M]

#include<stdio.h>
int main()
{
int a = 10;
switch(a)
{
case '1':
printf("ONE\n");
break;
case '2':
printf("TWO\n");
break;
defa1ut:
printf("NONE\n");
}
return 0;
}

If you think, this will print NONE, look carefully. The 'default' is
misspelled. I compiled this on gcc-2.95. Why doesnt the compiler
complain that this program is faulty? Why does it accept this program
as a valid one?

Thanks in advance,
Anjali.

[Dominik Zanettin]

On Mon, 05 Jul 2004 04:30:36 -0700, Anjali M wrote:

Why doesnt the compiler
complain that this program is faulty? Why does it accept this program
as a valid one?

because 'defa1ut:' is a valid lable. you could jump to it with goto.

regards
zhan

[Richard Bos]

g_********@rediffmail.com (Anjali M) wrote:

switch(a)
{
defa1ut:
printf("NONE\n");
} If you think, this will print NONE, look carefully. The 'default' is
misspelled. I compiled this on gcc-2.95. Why doesnt the compiler
complain that this program is faulty? Why does it accept this program
as a valid one?

Because it _is_ a valid program. defalut: is just another label. You
could even goto it, should you feel adventurous. Odd, perhaps, but true.

Richard

[Darrell Grainger]

On Mon, 5 Jul 2004, Anjali M wrote:

#include<stdio.h>
int main()
{
int a = 10;
switch(a)
{
case '1':
printf("ONE\n");
break;
case '2':
printf("TWO\n");
break;
defa1ut:
printf("NONE\n");
}
return 0;
}

If you think, this will print NONE, look carefully. The 'default' is
misspelled. I compiled this on gcc-2.95.
Why doesnt the compiler complain that this program is faulty?
If you have a valid symbol ending with a colon the compiler will assume
it is a label. I can put as many labels as I want in a program. They just
happen to look a little like a case statement, especially when you stick
it in the middle of a switch/case block. In other words, this program is
not faulty.

If you turned on warnings you might get the compiler warning you that you
have a label that you never used. It might even warn you that there is no
default case in the switch statement. But there is nothing in the C
standard indicating that it must do this.
Why does it accept this program as a valid one?
Because it sees the defalut: as a label to be used by a goto statement. It
is a valid C program. It is just not what you intended.
Thanks in advance,
Anjali.

--
Send e-mail to: darrell at cs dot toronto dot edu
Don't send e-mail to vi************@whitehouse.gov

[cedric]

Dominik Zanettin wrote:

because 'defa1ut:' is a valid lable.

Am I allowed to add this one to my collection of nice funny quotes ?

:-)

[Anjali M]

da*****@NOMORESPAMcs.utoronto.ca.com (Darrell Grainger) wrote in message news:<Pi*******************************@drj.pf>...

On Mon, 5 Jul 2004, Anjali M wrote:

#include<stdio.h>
int main()
{
int a = 10;
switch(a)
{
case '1':
printf("ONE\n");
break;
case '2':
printf("TWO\n");
break;
defa1ut:
printf("NONE\n");
}
return 0;
}

If you think, this will print NONE, look carefully. The 'default' is
misspelled. I compiled this on gcc-2.95.
Why doesnt the compiler complain that this program is faulty?

If you have a valid symbol ending with a colon the compiler will assume
it is a label. I can put as many labels as I want in a program. They just
happen to look a little like a case statement, especially when you stick
it in the middle of a switch/case block. In other words, this program is
not faulty.

If you turned on warnings you might get the compiler warning you that you
have a label that you never used. It might even warn you that there is no
default case in the switch statement. But there is nothing in the C
standard indicating that it must do this.

Why does it accept this program as a valid one?

Because it sees the defalut: as a label to be used by a goto statement. It
is a valid C program. It is just not what you intended.

Thanks in advance,
Anjali.

Okay, I get it now.
Thanks everybody,
Anjali

[Anjali M]

da*****@NOMORESPAMcs.utoronto.ca.com (Darrell Grainger) wrote in message news:<Pi*******************************@drj.pf>...

On Mon, 5 Jul 2004, Anjali M wrote:

#include<stdio.h>
int main()
{
int a = 10;
switch(a)
{
case '1':
printf("ONE\n");
break;
case '2':
printf("TWO\n");
break;
defa1ut:
printf("NONE\n");
}
return 0;
}

If you think, this will print NONE, look carefully. The 'default' is
misspelled. I compiled this on gcc-2.95.
Why doesnt the compiler complain that this program is faulty?

If you have a valid symbol ending with a colon the compiler will assume
it is a label. I can put as many labels as I want in a program. They just
happen to look a little like a case statement, especially when you stick
it in the middle of a switch/case block. In other words, this program is
not faulty.

If you turned on warnings you might get the compiler warning you that you
have a label that you never used. It might even warn you that there is no
default case in the switch statement. But there is nothing in the C
standard indicating that it must do this.

Why does it accept this program as a valid one?

Because it sees the defalut: as a label to be used by a goto statement. It
is a valid C program. It is just not what you intended.

Thanks in advance,
Anjali.

Okay, I get it now.
Thanks everybody,
Anjali

[Anjali M]

da*****@NOMORESPAMcs.utoronto.ca.com (Darrell Grainger) wrote in message news:<Pi*******************************@drj.pf>...

On Mon, 5 Jul 2004, Anjali M wrote:

#include<stdio.h>
int main()
{
int a = 10;
switch(a)
{
case '1':
printf("ONE\n");
break;
case '2':
printf("TWO\n");
break;
defa1ut:
printf("NONE\n");
}
return 0;
}

If you think, this will print NONE, look carefully. The 'default' is
misspelled. I compiled this on gcc-2.95.
Why doesnt the compiler complain that this program is faulty?

If you have a valid symbol ending with a colon the compiler will assume
it is a label. I can put as many labels as I want in a program. They just
happen to look a little like a case statement, especially when you stick
it in the middle of a switch/case block. In other words, this program is
not faulty.

If you turned on warnings you might get the compiler warning you that you
have a label that you never used. It might even warn you that there is no
default case in the switch statement. But there is nothing in the C
standard indicating that it must do this.

Why does it accept this program as a valid one?

Because it sees the defalut: as a label to be used by a goto statement. It
is a valid C program. It is just not what you intended.

Thanks in advance,
Anjali.

Okay, I get it now.
Thanks everybody,
Anjali