Performing MCDC for a condition (A && B && C) || D - unit-testing

Question

I have a situation where I need to write few test cases to get complete coverage of the code.
At a perticular branch I got following situation
if((A && B && C) || D)
if((A && B && C) || D)
{
//perform task 1'
}
else
{
//perform task 2;
}
But with this I am unable to get test case combination for MCDC..
What could be the way out here...

Answer 1

Your way out could be to use the tool MCDC to identify possible test cases.
Doing this manually could be really a lot of work, because of the many possible test pair combinations.
I am sorry that I cannot share all the details of my research. I would recomend to start with reading "An Investigation of Three Forms of the Modified Condition Decision Coverage (MCDC) Criterion", published by the FAA (DOT/FAA/AR-01/18).
Including Masking there are 52 possible test pairs. If you do a proper WhiteBox Analysis, including boolean short cut evaluation, then you will find even 63 MCDC test pairs. Apllying the set cover algorithm will result in 3 possible minimum test sets.
Doing some heuristics will lead to the maybe surprising test vector:
Test Pair for Condition 'a': 0 14 (Unique Cause)
Test Pair for Condition 'b': 8 14 (Unique Cause)
Test Pair for Condition 'c': 12 14 (Unique Cause)
Test Pair for Condition 'd': 0 1 (Unique Cause)
Testvector: Recommended Result: 0 1 8 12 14
0: a=0 b=0 c=0 d=0 (0)
1: a=0 b=0 c=0 d=1 (1)
8: a=1 b=0 c=0 d=0 (0)
12: a=1 b=1 c=0 d=0 (0)
14: a=1 b=1 c=1 d=0 (1)

Answer 2

With MC/DC you have to find, for each condition (input) i, a pair of combinations where only i toggles, and the output changes. For A to have a toggling effect on the result of (A && B && C) || D, B and C need to be true and D needs to be false. Similarly, for B to have a toggling effect, A and C need to be true, and D needs to be false. With D it is a bit different: The output toggles on D given that at least one of A, B, C is false. This gives the following combinations (x represents the toggling input):
A B C D
x 1 1 0 - for A
1 x 1 0 - for B
1 1 x 0 - for C
q r s x - for D, where q + r + s < 3
By multiplying out the lines for A, B, and C we get (added lines numbers and column R for the result):
# A B C D R
1 0 1 1 0 0 - for A
2 1 1 1 0 1 - for A
3 1 0 1 0 0 - for B
4 1 1 1 0 1 - for B (duplicate)
5 1 1 0 0 0 - for C
6 1 1 1 0 1 - for C (duplicate)
7 q r s x - for D, where q + r + s < 3
Lines 4 and 6 are duplicates of line 2. To represent line 7, we can use one of the lines 1,3,5 and add the same line with D set to 1, which gives the following result (picked line 1):
# A B C D R
1 0 1 1 0 0 - for A and D
2 1 1 1 0 1 - for A,B,C
3 1 0 1 0 0 - for B
4 1 1 0 0 0 - for C
5 0 1 1 1 1 - for D
That's one possible solution.

Answer 3

I think one way could be for all the false combination of (A && B && C) keep D = 1 and for true combination of (A && B && C) keep D = 0 and at the last Keep all inputs 0. So test case combination can be:
a b c d
0 1 1 1
1 0 1 1
1 1 0 1
1 1 1 0
0 0 0 0