Re: Declaring variables in "case" blocks?


"Keith Thompson" <kst-u@xxxxxxx> wrote:

"Robbie Hatley" <see.my.signature@xxxxxxxxxxxxxxxxxxxx> writes:
Greetings, group. I just found a weird problem in a program where
a variable declared in a {block} after a "case" keyword was being
treated as having value 0 even though its actual value should
have been something else. An extremely stripped-down version:

int Function (int something)
{
switch(something)
{
case WHATEVER:
{
int DumVar = 72;
// Some code uses DumVar. It seems to see the value
// of DumVar as being 0 ! What the heck???
break;
}
}
}

I think you've stripped down your code so far that you've eliminated
the error.

Maybe so.

The above is valid

That's what I was wondering about. Thanks.

and code that refers to DumVar will see its value as 72.

Theoretically, yes.

In practice (in my program, with my compiler), I was seeing 0,
which is why I was so puzzled.

You'd have been better off posting a small compilable program,
so you could confirm that the problem is still there.

I (mostly) did. If you #define WHATEVER, then call Function()
from main(), it will compile and run.

Hmmm... let me try that.

Ok, I just compiled and ran my stripped-down version here,
with a sprintf() of DumVar to a char buffer, and printed the buffer
in a popup message box. It displays as "72". Oops. Problem went
away. Weird.

I'm truly perplexed. I'll have to try to reproduce this bizarre
bug again.

But here's an example that illustrates the problem you're having:

#include <stdio.h>
int main(void)
{
int x = 42;
switch(x) {
int DumVar = 72;
case 42:
printf("ok, x = %d, DumVar = %d\n", x, DumVar);
break;
default:
printf("Huh?\n");
break;
}
return 0;
}

Nice try, but that's not my problem. I was careful to put
the variable declaration at the top of a {block}, just under a
CASE label, like so:

#include <stdio.h>
int main(void)
{
int x = 42;
switch(x) {
case 42:
{
int DumVar = 72;
printf("ok, x = %d, DumVar = %d\n", x, DumVar);
break;
}
default:
printf("Huh?\n");
break;
}
return 0;
}

and I was getting "DumVar = 0". Definitely wasn't a large
value (uninitialized RAM garbage) as you're getting here:

The output I get is:

ok, x = 42, DumVar = 1628438944

after a compiler warning:

c.c:6: warning: unreachable code at beginning of switch statement

The problem is that control jumps directly from the 'switch' to the
appropriate 'case' label, in this case to 'case 42:'. After that,
DumVar is visible, but you've skipped over its initialization.

Yep! I've seen that one bite even highly-experienced co-workers.
Since your declaration is never executed, your compiler declares
the variable as int for you, allocates some memory, but doesn't
initialize it to anything. So it's value could be anything from
-(2^31) to +(2^31)-1. (Assuming 32-bit int, 2's comp. rep.)

You can safely declare variables within a block following a case
label, but not within a block that's the entire body of the switch
statement.

Well, you could... but the code will never execute.

For more fun with switch statements, see question 20.35 in the
comp.lang.c FAQ, <http://www.c-faq.com/> (Duff's Device).

Whoa! It's actually legal to jump into the middle of a do loop
that way?! Cool machine!

--
Cheers,
Robbie Hatley
lone wolf aatt well dott com
triple-dubya dott Tustin Free Zone dott org


.

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