Question
What does Cyclomatic Complexity measure in software
engineering?Solution
Cyclomatic Complexity is a software metric used to measure the complexity of a program's control flow. Developed by Thomas McCabe in 1976, it quantifies the number of linearly independent paths through a program’s source code. This metric is important for several reasons: • Complexity Measurement: Cyclomatic Complexity provides an indication of the complexity of a program based on its control flow graph. The metric is calculated by counting the number of linearly independent paths through the program’s code. This helps in understanding how complex a module or function is, based on its decision points and branching. • Testing: A higher Cyclomatic Complexity value suggests more paths through the code, which means more test cases are needed to achieve thorough testing. Each independent path represents a unique test scenario that should be covered to ensure comprehensive testing of the module. • Maintainability: Programs with high Cyclomatic Complexity are generally harder to understand, maintain, and modify. By keeping Cyclomatic Complexity low, developers can create code that is easier to read, debug, and modify.
Statements: P ≤ Q < R; P ≥ S < T; S > U ≥ V
Conclusions:
I. S ≤ Q
II. T ≥ V
III. S > R
Statements: F ≥ G > H; I < J ≤ H; J > K
Conclusions:
I. F > J
II. K < G
III. H > K
Statements: A > C > W > S ≤ M ≥ N = T
Conclusion:
I. M > C
II. S > A
Statement: M < N; O ≥ P; O ≥ L; P > N ≥ Q
Conclusion:
I. N > L
II. L > Q
In this question, two statements are followed by two conclusions, numbered I and II. Find out which conclusion(s) is/are definitely true, based on the ...
Statements: C = A ≤ H < K ≥ L = Q; S = T ≥ K
Conclusion: I. C < T II. A = S
...Statements: F > V > W ≥ L > G; F ≤ O = M < I
Conclusions: I. M > LÂ Â Â II. V < I
Statement: A≤T<B =C ≤P<D;D>J ≥S
I. C >S
II. J < D
Which of the following set of symbols should be placed in the blanks respectively (from left to right) in the given expression in order to make the expr...
Statement: A < Z > I < B > S = O ≤ N
Conclusions:
I. S= N
II. B > O