Variable definition inside switch statement

Question

In the following code, why is the variable i not assigned the value 1?

#include       

int main(void)
{   
    int          


        

Answer 1

The initialization of variables with automatic storage durations is detailed in C11 6.2.4p6:

6. For such an object that does not have a variable length array type, its lifetime extends from entry into the block with which it is associated until execution of that block ends in any way. (Entering an enclosed block or calling a function suspends, but does not end, execution of the current block.) If the block is entered recursively, a new instance of the object is created each time. The initial value of the object is indeterminate. If an initialization is specified for the object, it is performed each time the declaration or compound literal is reached in the execution of the block; otherwise, the value becomes indeterminate each time the declaration is reached.

I.e. the lifetime of i in

switch(a) {
    int i = 2;
    case 1: printf("%d",i);
            break;
    default: printf("Hello\n");
}

is from { to }. Its value is indeterminate, unless the declaration int i = 2; is reached in the execution of the block. Since the declaration is before any case label, the declaration cannot be ever reached, since the switch jumps to the corresponding case label - and over the initialization.

Therefore i remains uninitialized. And since it does, and since it has its address never taken, the use of the uninitialized value to undefined behaviour C11 6.3.2.1p2:

2. [...] If the lvalue designates an object of automatic storage duration that could have been declared with the register storage class (never had its address taken), and that object is uninitialized (not declared with an initializer and no assignment to it has been performed prior to use), the behavior is undefined.

(Notice that the standard itself here words the contents in the clarifying parenthesis incorrectly - it is declared with an initializer but the initializer is not executed).