Ques1. What is the output of the following ?
Options :
A. 3, 2, 15
B. 2, 3, 20
C. 2, 1, 15
D. 1, 2, 5
Answer : A
Explanation : >> int a[5] = {5, 1, 15, 20, 25}; The variable arr is declared as an integer array with a size of 5 and it is initialized to
a[0] = 5, a[1] = 1, a[2] = 15, a[3] = 20, a[4] = 25.
>> int i, j, m; The variable i, j, m are declared as an integer type.
>> i = ++a[1]; becomes i = ++1; Hence i = 2 and a[1] = 2
>> j = a[1]++; becomes j = 2++; Hence j = 2 and a[1] = 3.
>> m = a[i++]; becomes m = a[2]; Hence m = 15 and i is incremented by 1(i++ means 2++ so i=3)
>> printf("%d, %d, %d", i, j, m); It prints the value of the variables i, j, m
Ques2. What is the output of the following ?
Options :
A. Garbage Value
B. 5
C. 6
D. 0
Answer : D
Explanation : When an array is partially initialized at the time of declaration then the remaining elements of the array is initialized to 0 by default.
Ques3. What is the output of the following?
Assume that the array begins at 65472 and each integer occupies 2 bytes
Options :
A. 65474, 65488
B. 65480, 65488
C. 65480, 65496
D. 65474, 65476
Answer : C
Explanation : The base address(also the address of the first element) of array is 65472.
For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore, a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 + (4 ints * 2 bytes) = 65480
Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes".
Hence, beginning address 65472 + 24 = 65496. So, &a+1 = 65496
Ques4. What is the output of the following ?
Options :
A. for Geeks
B. Geeks
C. compile time error
D. no output
Answer : A
Explanation : It skips the 6 characters and prints the given string.
Ques5. What is the output of the following ?
