Dive into the complexities of C programming with the 'Hardest C Programming Trivia Quiz!' Explore various coding challenges, from Fibonacci sequence generation to memory address comparisons, and enhance your understanding of C language fundamentals.
Counting microseconds
Counting the number of key operations
Counting the number of statements
Counting the kilobytes of algorithm
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Best case
Worst case
Average case
Null case
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Item is somewhere in the middle of the array
Item is not in the array at all
Item is the last element in the array
Item is the last element in the array or is not there at all
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For relatively permanent collections of data
for the size of the structure and the data in the structure are constantly changing
For both of above situation
For none of above situation
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Are easier to develop than single programming systems
Execute each job faster
Execute more jobs in the same time period
Are used only one large mainframe computers.
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Abaabaaabaa
Aaaabaaaa
Baaaaabaaaab
Baaaaabaa
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Int and float
Extern
Both 1 & 2
Neither 1 nor 2
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Main(int argc, char argv)
Main(char argv[ ], int *argc)
Main(char* argv[ ],int argc)
Main(int argc,char* argv[ ])
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# define pie = 22/7
#define pie 22/7
#define pie= 3.142
# Define pie 22/7
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0
Run-time error
Undefined
Compilation error
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Prints numbers from 0 to n-1
Prints numbers from n-1 to 0
Prints first n Fibonacci numbers
Prints first n Fibonacci numbers in reverse order.
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13.14
13
13.00
None of the above
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5 5 5
5 5 garbage value
5 5 0
Compile Error
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X is 97
X is 98
X is 99
Compilation Error
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Different addresses are printed
Same address is printed separated by a space
Compilation error
Nothing
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Root is visited before right subtree
Left subtree is always visited before right subtree
Root is visited after left subtree
All of the above
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Compile Error
It’ll always print Equal.
It’ll always print Not Equal.
Since elements of arrPtr aren’t initialized in the program, it’ll print either Equal or Not Equal.
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22
21
01
02
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Not Possible
Yes, possible by storing XOR of addresses of previous and next nodes.
Yes, possible by storing XNOR of addresses of previous and next nodes.
Yes, possible by storing XOR of the current node and previous node
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10, 20, 15, 23, 25, 35, 42, 39, 30
15, 10, 25, 23, 20, 42, 35, 39, 30
15, 20, 10, 23, 25, 42, 35, 39, 30
15, 10, 23, 25, 20, 35, 42, 39, 30
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Void *
A pointer of allocated memory type
Void **
Int *
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Even
Odd
Default
Compile-time error
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22 bytes
14 bytes
18 bytes
10 bytes
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O(n)
O(n log(n))
O(1)
O(log n)
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1
2
6
Runtime Error
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-2
-1
1
2
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2
3
4
5
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The cost of searching an AVL tree is θ(log n) but that of a binary search tree is O(n)
The cost of searching an AVL tree is θ(log n) but that of a complete binary tree is θ(n log n)
The cost of searching a binary search tree is O(log n ) but that of an AVL tree is θ(n)
The cost of searching an AVL tree is θ(n log n) but that of a binary search tree is O(n)
4040
4050
5040
5050
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Compilation error
Gnirt
String
Gnirts
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Register
Cache
Main memory
Disk
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5
15
40
55
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Feof()
EOF
Both feof() and EOF
None of the mentioned
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Int
Float
Void
Error
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O(n)
O(n log(n))
O(log n)
O(n^2)
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Strlen()
Strcmp()
Strcat()
Strcpy()
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List
Stack
Queue
None
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Aba*b*bba
Ab(ab)*bba
Ab(a+b)*bba
All of the mentioned
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Lottery scheduling
FCFS algorithm
Shortest Job Next
None of the above
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