Inline ternary operator handling in calculation of NPath complexity

[Andrey Godyaev]

I have encountered difficulty in calculating of NPath complexity for methods containing inline ternary operators in complex expressions.
Here is an example of calculations and i need comfirmation i've done them correctly:
https://gist.github.com/attatrol/62f017acae2cff99fe1c

[Roman Ivanov]

Hi Andrey,

>     //NP(ternary) = 2

    int i = true ? 1 : 2;

Correct.

from Npath article:

Syntax: (exprl) ? (expr2):(expr3)

NP(?)=NP((expr1))+NP((expr2)) +NP((expr3))+2.

NP(ternary) = 0 + 0 + 0 + 2 = 0

NP((expr1)) = 0 - because there are no condition operators. In article there is example that prove this.

>     //NP(cycle) = 2

    for(; i<10; i++) {

    }

Correct.

from Npath article:

Syntax: for ((exprl); (expr2); (expr3)) (for-range)

NP(for)= NP((for-range)) + NP((expr1)) + NP((expr2)) + NP((expr3)) + 1

NP(for) = 1 + 0 + 0 + 1

>     //NP(cycle) = NP(exprs) + NP(range) + 1 = 2 + 1 + 1 = 4

    for(int i = true ? 1 : 2; i < 10; i++) {

    }

Correct.

NP(for)= NP((for-range)) + NP((expr1)) + NP((expr2)) + NP((expr3)) + 1

NP(for) = 1 + 2 + 0 + 0 + 1 = 4

>    //NP(cycle) = NP(expr) + NP(range) + 1 = 2 + 1 + 1 = 4

    for(int i = 1; i < true ? 10 : 15; i++) {

           //NP(range) = 1

    }

Correct.

NP(for)= NP((for-range)) + NP((expr1)) + NP((expr2)) + NP((expr3)) + 1

NP(for) = 1 + 0 + 2 + 0 + 1 = 4

>     //NP(cycle) = NP(expr) + NP(range) + 1 = 0 + 2 + 1 = 3

    for(int i = 1; i < j; i++) {

        //NP(range) = NP(ternary) = 2

        int j = true ? 10 : 15;

    }

Correct.

NP(for)= NP((for-range)) + NP((expr1)) + NP((expr2)) + NP((expr3)) + 1

NP(for) = 2 + 0 + 0 + 0 + 1 = 3

thanks,

Roman Ivanov

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