what will be the cyclomatic complexity of this c++ code

Question

int x = 99;
int a = 4;
int parity;
while (x < 100)
{
    if (a % 2 == 0)
    {
        parity = 0;
    }
    else
    {
        parity = 1;
    }
    switch (parity)
    {
    case 0:
        cout << "even ";
    case 1:
        cout << " odd ";
    default:
        cout << "Error";
    }
    x++;
}
int p = 1;

as the parity value hass to pass through if else statement so will there be aby independent path in which only the default case will be executed cuz when parity is 0 case:0,1and default will execute dute to fall through similarly with case1 so what will be the cyclomatic complexity (acc to me it will be 4)

Answer 1

The cyclomatic complexity of the given code is 4. This is because the code contains 4 independent paths of execution. These paths are determined by the conditional statements in the code, including the if statement and the switch statement. Each of these statements creates a new independent path, and the combination of these paths determines the overall cyclomatic complexity of the code.

In this case, the if statement creates two independent paths, one for when a % 2 == 0 and one for when a % 2 != 0. The switch statement also creates two independent paths, one for each of the case statements. Together, these paths create a total of 4 independent paths of execution.

It is worth noting that the default case in the switch statement will not create a new independent path, as it will only be executed if none of the other case statements is executed. In other words, the default case is not an independent path of execution, but rather a fallback option that is only executed if the other paths fail.

EDIT

These are 4 distinct paths can be found from the code:

• The first path is when parity is 0 and x is less than 100. In this case, the code will print "even", and then increment x by 1.

• The second path is when parity is 1 and x is less than 100. In this case, the code will print "odd", and then increment x by 1.

• The third path is when parity is 0 and x is not less than 100. In this case, the code will not execute the while loop, and will not print anything.

• The fourth path is when parity is 1 and x is not less than 100. This is the same as the third path, and the code will not execute the while loop or print anything.