UGC NET Computer Science C Programming Language Previous Year Questions (PYQs)
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int i, j;
for(i = 1; i < 5; i += 2)
for(j = 1; j < i; j += 2)
printf("%d", j);
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Solution
For i = 1, inner loop condition j < i → 1 < 1 → false → no output.
For i = 3, inner loop runs with j = 1, prints 1, then j = 3 → condition fails.
Hence, final output is 1.
Consider the operators used in C programming given below:
A. &&
B. +=
C. >>
D. >=
E. ?:
Choose among the following the correct order of precedence of the operators given above (higher to lower):
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Solution
Operator precedence (highest → lowest) is:
>> (shift) → >= (relational) → ?: (ternary) → += (assignment) → && (logical AND).
void main() {
int *i, a = 12, b = 2, c;
c = (a = a + b, b = a / b, a = a * b, b = a - b);
i = &c;
printf("%d", -(*i));
}
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Solution
Step-by-step explanation:
a = 12, b = 2
a = a + b → a = 14
b = a / b → b = 7
a = a * b → a = 98
b = a - b → b = 91
→ c = 91
Hence, -(*i) = -91
Output: 91 (negative of c printed as -(-91) → 91)
int x = 128, y = 110;
do {
if (x > y) x = x - y;
else y = y - x;
} while (x != y);
printf("%d", x);
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Solution
Repeated subtraction implements Euclid’s algorithm; the loop ends with x = y = gcd(128,110) = 2.
Only legal pointer operations:
A. pointer + number → pointer
B. pointer – number → number
C. pointer + pointer → pointer
D. pointer – pointer → pointer
E. pointer – pointer → number
Choose the most appropriate answer from the options given below:
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Solution
pointer + number → pointer (A)
pointer - pointer → number (E)
What would be the equivalent pointer expression for referring the array element ar[m][n][o]?
- *(*(*(ar) + m + n) + o)
- (*(*(*(ar + m) + n) + o))
- (*(*(ar + m) + n) + o)
- *(*(*(ar + m) + n) + o)
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Solution
Consider ar as a 3-dimensional array declared as:
data_type ar[a][b][c];
Here, the array element ar[m][n][o] can be represented in pointer form as:
*(*(*(ar + m) + n) + o)Step-by-step understanding:
aris a pointer to the first 2D arrayar[0].ar + mmoves the pointer to the mth 2D array →ar[m].*(ar + m)gives the address of the first 1D array insidear[m].*(ar + m) + nmoves to the nth 1D array →ar[m][n].*(*(ar + m) + n)gives the address of the first element of that 1D array.*(*(*(ar + m) + n) + o)finally gives the elementar[m][n][o].
✅ Final Answer: *(*(*(ar + m) + n) + o)
UGC NET Computer Science
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UGC NET Computer Science
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and More.
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