ISTQB Foundation Level CTFL-Foundation New Questions

The impossibility of complete testing has several important consequences. It means we can never be certain that the program is entirely bug-free (A). There is no definite stopping point for testing, which can make it easier for some managers to justify minimal testing (B). Additionally, there is no simple answer for which testing tasks should always be required, as each task takes time that could be allocated to other critical activities (C). Therefore, all these consequences (D) are valid and important considerations in software testing. Reference: ISTQB CTFL Syllabus V4.0, Section 2.2​​.

When a defect is detected and fixed, the software should be retested to confirm that the original defect has been successfully removed. This process is called confirmation testing (or re-testing). According to the ISTQB CTFL Syllabus, version 4.0, confirmation testing involves re-running a test to verify that a specific defect has been fixed. The main objective of this type of testing is to ensure that the initial defect no longer exists after the fix has been applied.

The syllabus states that confirmation testing is distinct from regression testing, which checks for unintended side effects of changes in the software. Confirmation testing specifically targets the areas where defects were found and fixed to ensure those defects are resolved​​.

The concept of introducing a small change to the program and having the effects of that change show up in some test is known as "introducing mutations." This is part of mutation testing, where small changes (mutations) are made to the code to see if the existing test cases can detect the errors introduced by these changes. The goal is to evaluate the effectiveness of the test cases in identifying defects. Reference: ISTQB CTFL Syllabus V4.0, Section 4.3.4​​.

McCabe’s Cyclomatic Complexity is a metric used to measure the complexity of a program. It is calculated using the formula:

Cyclomatic Complexity=E−N+2P\text{Cyclomatic Complexity} = E - N + 2PCyclomatic Complexity=E−N+2P

where:

E = number of edges in the flow graph

N = number of nodes in the flow graph

P = number of connected components (usually 1 for a single program)

In the given program, there are 2 decision points (IF and THEN). The cyclomatic complexity is calculated as:

Cyclomatic Complexity=2−1+2=3\text{Cyclomatic Complexity} = 2 - 1 + 2 = 3Cyclomatic Complexity=2−1+2=3

Hence, the answer is B. Reference: ISTQB CTFL Syllabus V4.0, Section 4.3.2​​.