Consider the following C program segment. while(first last) notPresent = TRUE; The cyclomatic complexity of the program segment is …
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GATE CSE 2015 Set 1 · Q42
NAT2 marksEasy
Time
Consider the following C program segment.while(first <= last)
{
if (array[middle] < search)
first = middle + 1;
else if (array[middle] == search)
found = TRUE;
else last = middle - 1;
middle = (first + last)/2;
}
if (first > last) notPresent = TRUE;The cyclomatic complexity of the program segment is _________.
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Solution
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Official answer: 5
Programming in CCyclomatic complexity of a control flow graph can be computed using the number of predicate (decision) conditions as:
Identifying the decision/predicate conditions in the given program segment:
while(first <= last)— 1 predicate conditionif (array[middle] < search)— 1 predicate conditionelse if (array[middle] == search)— 1 predicate conditionif (first > last)— 1 predicate condition
Total number of predicate conditions:
Therefore, the cyclomatic complexity is:
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Examiner trap
The misconception this question tests
Cyclomatic complexity of a program can be determined as , where is the total count of decision/conditional statements.
This Programming in C previous year question is Q42 of the GATE CSE 2015 Set 1 paper. Find more GATE CSE PYQs on Programming in C in Programming and Data Structures, or browse all GATE CSE PYQs.
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Programming in C in GATE CSE
Programming in C has 74 GATE CSE PYQs across 25 papers from 2011 to 2026. 17 of them are 2-mark NAT questions like this one, and GATE CSE 2015 Set 1 has 4.
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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.