Cyclomatic Complexity of switch case statement

Question

I'm confusing about the CC of switch statement

If I have following code:

if (n >= 0) {
    switch(n) {
        case 0:
        case 1: 
            printf("zero or one\n");
            break;
        case 2: 
            printf("two\n");
            break;
        case 3:
        case 4: 
            printf("three or four\n");
            break;
        }
    }
else {
    printf ("negative\n");
}

what is the CC?

I found a post said that it's 5, with this diagram

(the edges are 17, not 16, I think it's a typo)

It says that we only need to count case 0 and case 1 as one

But I think the diagram should be:

Edges: 17,
Nodes: 13,
17 - 13 + 2P = 6

I count every cases as 1

My OOSE professor said it's 6, but in different way

He said:

init     => 1  
if       => 1  
switch   => 1  
case 0 1 => 1  
case 2   => 1  
case 3 4 => 1

so it should be 6

What's the correct answer?
I'm really confused, thanks.

---

edited:
Now I think it's 7. yes, 7
Because if n is more than 5, will just do nothing and exit the switch statement.

then we get this diagram:

now E = 18
18 - 13 + 2 = 7

am I correct..?
really, really, really confused...

Answer 1

Code metric tools I've worked with count each case as a separate branch, even if it's a fall through case.

But this is an arbitrary choice. Code metric tools tend to err on the side of caution by default. The way the switch statement will ultimately be evaluated is an internal implementation detail that will vary based on the type of input and number of cases (at least in C#).

The go-to answer for reducing cyclomatic complexity caused by switch statements is to convert the cases/outputs into a dictionary. In your example, it would be something like the code sample below. Be aware that this is only for readability/maintainability. If your switch statement is long enough the .Net compiler will automatically convert it to a dictionary for you, so there is no performance gain.

var outputs = new Dictionary<int, string>()
            {
                { 0, "zero or one\n" },
                { 1, "zero or one\n" },
                { 2, "two\n" },
                { 3, "three or four\n" },
                { 4, "three or four\n" }
            };

if (n >= 0)
{
    printf(outputs[n]);
}

Answer 2

Ok, I have found the answer.

from McCabe.com, page 26 and 27

The answer is 5, because the original version of CC by McCabe counts a fall-through case as 1.