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Q.1 Medium
What will be the output of the following C code?

int main() {
int x = 5;
int *p = &x;
int **q = &p;
printf("%d", **q + 1);
return 0;
}
A 6
B 5
C Compilation error
D Garbage value
Correct Answer:  A. 6
Explanation:

The correct answer is 6 because q is a pointer to pointer that points to p, which points to x. When we dereference q, we get the value of x (which is 5), and then adding 1 gives us 6, which is printed. The key concept here is understanding pointer dereferencing: a single asterisk dereferences one level, so double asterisks dereference two levels, bringing us from q to p to the actual value stored in x. Any other answer would be incorrect because it would either misunderstand the double dereferencing operation or incorrectly interpret what value is being accessed—for instance, if someone only dereferenced once or forgot to add 1, they would get a different result.

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Q.2 Medium
What is the output of the following code snippet?

int main() {
char arr[] = "GATE";
char *p = arr;
printf("%d\n", sizeof(arr));
printf("%d", sizeof(p));
return 0;
}
A 4 and 4
B 5 and 4
C 5 and 8 (on 64-bit system)
D 4 and 8 (on 64-bit system)
Correct Answer:  C. 5 and 8 (on 64-bit system)
Explanation:

The output is 5 and 8 because sizeof(arr) returns 5 bytes (4 characters plus 1 null terminator in the string "GATE"), while sizeof(p) returns 8 bytes on a 64-bit system since p is a pointer variable and pointers occupy 8 bytes in 64-bit architecture. The key concept is that sizeof() behaves differently for arrays versus pointers: when applied to an array, it returns the total memory allocated for the entire array, but when applied to a pointer, it only returns the size of the pointer itself, not what it points to. Other options would be incorrect because they either miscalculate the array size by forgetting the null terminator, use incorrect pointer sizes (like 4 bytes for 32-bit systems), or confuse the size of the array with the size of the pointer variable.

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Q.3 Medium
Which of the following operations has an average time complexity of O(1) in a Hash Table with good hash function and low collision rate?
A Insertion only
B Search only
C Both insertion and search
D Deletion only
Correct Answer:  C. Both insertion and search
Explanation:

Hash Tables achieve O(1) average-case time complexity for both insertion and search operations when using a well-designed hash function that minimizes collisions and maintains a good load factor.

Deletion also operates in O(1) average time.

However, in worst-case scenarios with poor hash functions or high collision rates, these operations can degrade to O(n).

The key to performance is maintaining low collision rates through techniques like chaining or open addressing.

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Q.4 Medium
Which data structure is most suitable for implementing a priority queue where elements with higher priority need to be dequeued first?
A Array
B Linked List
C Min-Heap or Max-Heap
D Graph
Correct Answer:  C. Min-Heap or Max-Heap
Explanation:

A Heap (either Min-Heap or Max-Heap) is the optimal data structure for implementing priority queues with O(log n) insertion and deletion time complexity.

In a Max-Heap, the element with the highest priority is always at the root, enabling efficient extraction of the maximum element.

Arrays and Linked Lists would require O(n) time for priority-based operations, while Graphs are unsuitable for this purpose.

Heaps are fundamental in algorithms like Dijkstra's and Prim's for finding shortest paths.

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Q.5 Medium C Programming
What is the output of the following C code?
int x = 5;
int y = ++x + x++;
printf("%d", y);
A 11
B 12
C 13
D Undefined behavior
Correct Answer:  D. Undefined behavior
Explanation:

This code exhibits undefined behavior because x is modified twice between sequence points without an intervening sequence point. The result depends on the compiler's implementation.

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Q.6 Medium C Programming
What is the difference between scanf() and gets() functions?
A scanf() can read formatted input while gets() reads only strings
B gets() is safer than scanf()
C There is no difference
D scanf() stops at whitespace while gets() doesn't
Correct Answer:  A. scanf() can read formatted input while gets() reads only strings
Explanation:

scanf() reads formatted input based on format specifiers and stops at whitespace. gets() reads a string until a newline is encountered. Note: gets() is deprecated due to buffer overflow vulnerabilities.

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Q.7 Medium C Programming
What is the correct way to allocate memory for a single integer using malloc()?
A int *ptr = (int *)malloc(sizeof(int));
B int *ptr = malloc(sizeof(int));
C int *ptr = (int *)malloc(4);
D Both A and B
Correct Answer:  D. Both A and B
Explanation:

Both A and B are correct. Option A uses explicit type casting which is optional in C (not in C++). Option B avoids casting. Using sizeof(int) is preferred over hardcoding 4, as int size may vary across systems.

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Q.8 Medium C Programming
What is the output of the following C code?
int x = 5;
int *ptr = &x;
printf("%d %d", *ptr, x);
A 5 5
B Address Address
C Garbage value 5
D Compilation error
Correct Answer:  A. 5 5
Explanation:

*ptr dereferences the pointer to access the value at the address it points to, which is x = 5. So both *ptr and x print 5.

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Q.9 Medium C Programming
What happens when you use the strcpy() function without bounds checking?
A It prevents buffer overflow automatically
B It may cause buffer overflow if source string is longer than destination
C It returns an error code
D It truncates the source string automatically
Correct Answer:  B. It may cause buffer overflow if source string is longer than destination
Explanation:

strcpy() does not perform bounds checking. If the source string is longer than the destination buffer, it will write beyond the buffer boundary, causing a buffer overflow. This is a security vulnerability. Using strncpy() is safer.

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Q.10 Medium C Programming
What is the output of the following C code?
int arr[] = {10, 20, 30};
int *ptr = arr;
printf("%d", *(ptr + 1));
A 10
B 20
C 30
D Address of second element
Correct Answer:  B. 20
Explanation:

ptr points to arr[0]. ptr + 1 points to arr[1]. *(ptr + 1) dereferences to get the value at arr[1] which is 20. Pointer arithmetic adds sizeof(int) to the address for each increment.

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