Question

In the following code:

int i = 0;

switch(i)
{
    case 0:
        cout << "In 0" << endl;
        i = 1;
        break;
    case 1:
        cout << "In 1" << endl;
        break;
}

What will happen? Will it invoke undefined behavior?

Answer 1

No undefined behavior. But the value of i is only tested when the code reaches switch (i). So case 1: will be skipped (by the break; statement).

The switch keyword does not mean “run code whenever the value of i is 0 / 1″. It means, check what i is RIGHT NOW and run code based on that. It doesn’t care what happens to i in the future.

In fact, it’s sometimes useful to do:

for( step = 0; !cancelled; ++step ) {
    switch (step)
    {
        case 0:
            //start processing;
            break;

        case 1:
            // more processing;
            break;

        case 19:
            // all done
            return;
    }
}

And changing the control variable inside a case block is extremely common when building a finite state machine (although not required, because you could set next_state inside the case, and do the assignment state = next_state afterward).