Output of C programs | Set 38

Ques1. What is the output of the following ?

#include <stdio.h>
void main()
{
    int a[5] = { 5, 1, 15, 20, 25 };
    int i, j, m;
    i = ++a[1];
    j = a[1]++;
    m = a[i++];
    printf("%d, %d, %d", i, j, m);
}

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 ?

void main()
{
    int arr[10] = { 1, 2, 3, 4, 5 };
    printf("%d", arr[5]);
}

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

void main()
{
    int a[3][4] = { 1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0 };
    printf("%u, %u", a + 1, &a + 1);
}

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 ?

#include <stdio.h>
void main()
{
    printf(6 + "Geeks for Geeks");
}

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 ?

#include <stdio.h>
void main()
{
    char* s = "hello";
    char* p = s;
    printf("%p\t%p", p, s);
}

Options :
A. Different address is printed
B. Same address is printed
C. Run time error
D. Nothing

Answer : B

Explanation : *s will hold the value “hello”and that string is assigned to *p which will start pointing to same physical location.

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