Declared variable within a switch block

Question

#include <stdio.h>

int main()
{
        int expr = 0;

        switch (expr)
        {
                int i;

                case 0:
                        i = 17;
                default:
                        printf("%d\n", i);
        }
        return 0;
}

In the above code, variable i declared within the switch case block and print into default case. Use following command to compiled in GCC compiler on linux platform.

gcc -Wall switch.c

and it is print correct output.

So, is it defined behavior or undefined behavior?

It is UB because it will access unintialized value if expr is not 0. But in this case, no — Ajay Brahmakshatriya, Aug 30 '17 at 09:16
OK, forgot to ask, why did you think this would be undefined behaviour? Reason? — Sourav Ghosh, Aug 30 '17 at 09:21
Using a variable before it has been initialized is undefined behaviour. The fact that it has been declared in the switch block rather than outside the switch block doesn't matter. — Jabberwocky, Aug 30 '17 at 09:22
@SouravGhosh I thought inside the switch statement, without any casses declared variable not execute. — Jayesh, Aug 30 '17 at 09:23
See [6.8.4.2 The switch statement p7](http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf) — BLUEPIXY, Aug 30 '17 at 09:40
@BLUEPIXY Ah, was just updating the same, to add to the answer. — Sourav Ghosh, Aug 30 '17 at 09:44

Sourav Ghosh · Answer 1 · 2017-08-30T09:42:06.617

In this case, it is not undefined behaviour.

For case 0 (when expr == 0, which is your case), i gets assigned a value before being used (value being read from).

OK, to elaborate a bit more, for the snippet

switch (expr)
    {
            int i;

            case 0:
                    i = 17;
            default:
                    printf("%d\n", i);
    }

just makes the variable i defined in the block scope. Even if, the code would have been written as

            int i = 0; //or any value

the value of i not initialized, it's just the identifier is visible in the scope. You must have another statement assigning value to i before you can make use of it.

In this regard, the C11 standard has a very enlightening example and description. let me quote it, from chapter §6.8.4.2/P7

EXAMPLE In the artificial program fragment
switch (expr)
{
int i = 4;
f(i);
case 0:
i = 17;
/* falls through into default code */
default:
printf("%d\n", i);
}
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. [....]

Declared variable within a switch block

1 Answers1