Question 1

What will be the output of the program if the array begins at address 65486?

#include<stdio.h>

int main()
{
    int arr[] = {12, 14, 15, 23, 45};
    printf("%u, %u\n", arr, &arr);
    return 0;
}

Answer

Answer: B

Step 1: int arr[] = {12, 14, 15, 23, 45}; The variable arr is declared as an integer array and initialized.

Step 2: printf("%u, %u\n", arr, &arr); Here,

The base address of the array is 65486.

=> arr, &arr is pointing to the base address of the array arr.

Hence the output of the program is 65486, 65486

Question 2

What will be the output of the program ?

#include<stdio.h>

int main()
{
    int arr[1]={10};
    printf("%d\n", 0[arr]);
    return 0;
}

Answer

Answer: B

Step 1: int arr[1]={10}; The variable arr[1] is declared as an integer array with size '2' and it's first element is initialized to value '10'(means arr[0]=10)

Step 2: printf("%d\n", 0[arr]); It prints the first element value of the variable arr.

Hence the output of the program is 10.

Question 3

What will be the output of the program in Turb C (under DOS)?

#include<stdio.h>

int main()
{
    int arr[5], i=0;
    while(i<5)
        arr[i]=++i;

    for(i=0; i<5; i++)
        printf("%d, ", arr[i]);

    return 0;
}

Answer

Answer: B

Since C is a compiler dependent language, it may give different outputs at different platforms. We have given the TurboC Compiler (Windows) output.

Please try the above programs in Windows (Turbo-C Compiler) and Linux (GCC Compiler), you will understand the difference better.

Question 4

What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?

#include<stdio.h>

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

Answer

Answer: B

Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a[3][4] is declared as an integer array having the 3 rows and 4 colums dimensions.

Step 2: printf("%u, %u\n", a+1, &a+1);

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, begining address 65472 + 24 = 65496. So, &a+1 = 65496

Hence the output of the program is 65480, 65496

Question 5

What will be the output of the program ?

#include<stdio.h>

int main()
{
    static int arr[] = {0, 1, 2, 3, 4};
    int *p[] = {arr, arr+1, arr+2, arr+3, arr+4};
    int **ptr=p;
    ptr++;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    *ptr++;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    *++ptr;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    ++*ptr;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    return 0;
}

Answer

Answer: C

No answer description is available. Let's discuss.

Questions for: Arrays