CODEWITHSUNDEEP
c++switch-statement
Daniel
Daniel

Reputation: 31579

Changing switch variable inside a case

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?

Upvotes: 17

Views: 12680

Answers (4)

Bebu
Bebu

Reputation: 65

Your output would be :

"In 0"

even if you assign the value i = 1 it wont be reflected because switch does not operate in iteration, it is one time selection as break would make it go out of the switch statement.

Upvotes: 0

Ben Voigt
Ben Voigt

Reputation: 283763

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).

Upvotes: 23

CB Bailey
CB Bailey

Reputation: 792677

There's no issue here. The expression in the switch condition is evaluated when it is reached. It doesn't have to be a variable and if it is the variable can be subsequently modified without any effect on the behaviour of the switch statement.

Upvotes: 2

Joe
Joe

Reputation: 57179

You break out of this switch statement after you set it to 1 which is defined behavior so it will never enter case 1.

Upvotes: 2

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