Data Structure MCQ Exam Quiz!

2.

What will be the output of the program ? #include void fun(int **p); int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; int *ptr; ptr = &a[0][0]; fun(&ptr); return 0; } void fun(int p) { printf("%d", p); }

1
2
3
4

Correct Answer

A. 1

Explanation

The program declares a 2D array 'a' with 3 rows and 4 columns. It then declares a pointer 'ptr' and assigns it the address of the first element of the array. The function 'fun' takes a double pointer as a parameter. In the main function, the address of 'ptr' is passed to 'fun'. Inside 'fun', the value at the address pointed by the double pointer is printed, which is the first element of the array 'a'. Therefore, the output of the program will be 1.

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3.

What does the following declaration mean? int (ptr)[10]*

Ptr is array of pointers to 10 integers
Ptr is a pointer to an array of 10 integers
Ptr is an array of 10 integers
Ptr is an pointer to array

Correct Answer

A. Ptr is a pointer to an array of 10 integers

Explanation

The declaration "int (*ptr)[10]" means that "ptr" is a pointer to an array of 10 integers.

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4.

What is the output of the code given below? #include void main() { float arr[5] = {12.5, 10.0, 13.5, 90.5, 0.5}; float ptr1 = &arr[0]; float ptr2 = ptr1 + 3; printf("%f ", *ptr2); printf("%d", ptr2 - ptr1);}

90.500000 3
90.500000 12
10.000000 12
0.500000 3

Correct Answer

A. 90.500000 3

Explanation

The code initializes an array of float values and assigns the memory address of the first element to ptr1. Then, ptr2 is assigned the memory address of the element at index 3 (90.5).

In the first printf statement, *ptr2 is used to print the value at the memory address pointed by ptr2, which is 90.5.

In the second printf statement, ptr2 - ptr1 is used to calculate the difference between the memory addresses pointed by ptr2 and ptr1. Since ptr2 is pointing to the element at index 3 and ptr1 is pointing to the element at index 0, the difference is 3.

Therefore, the output of the code is "90.500000 3".

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5.

What is the output of the code given below? #include void main() { int arr[] = {10, 20, 30, 40, 50, 60}; int ptr1 = arr; int ptr2 = arr + 6; printf("Number of Elements : %d", (ptr2 - ptr1)); printf("Number of Bytes : %d", (char)ptr2 - (char) ptr1); }

Number of Elements : 5 Number of Bytes: 20
Number of Elements : 6 Number of Bytes: 24
Compile Time Error
Run Time Error

Correct Answer

A. Number of Elements : 6 Number of Bytes: 24

Explanation

The code declares an array "arr" with 6 elements. Two pointers "ptr1" and "ptr2" are assigned the address of the first element and the address of the element at index 6 respectively. The expression "(ptr2 - ptr1)" calculates the number of elements between the two pointers, which is 6. The expression "(char*)ptr2 - (char*)ptr1" calculates the number of bytes between the two pointers, which is 24. Therefore, the output of the code is "Number of Elements : 6 Number of Bytes: 24".

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6.

What is the output of the code given below? # include void fun(int ptr) { ptr = 30; } int main() { int y = 20; fun(&y); printf("%d", y); }

20
30
Compile Time error
Run Time error

Correct Answer

A. 30

Explanation

The code defines a function called "fun" that takes a pointer to an integer as a parameter. Inside the function, the value at the memory location pointed to by the pointer is changed to 30.

In the main function, an integer variable y is declared and initialized to 20. Then, the function "fun" is called with the address of y as the argument. This means that the function will modify the value of y.

Finally, the value of y is printed, which is now 30 because it was modified by the "fun" function.

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7.

Consider a two dimensional array int A[100][100]; // assume that an integer requires 4 bytes. Starting address of the array is 5000 What is the address of the element A[4][5] if the allocation is row major ?

Address cannot be determined
5080
6620
6120

Correct Answer

A. 6620

Explanation

In a two-dimensional array allocated in row major order, the elements are stored in consecutive memory locations row by row. Since each integer requires 4 bytes, the address of A[4][5] can be calculated as follows:

Starting address of the array = 5000
Size of each row = 100 * 4 = 400
Size of previous rows = 4 * 400 = 1600
Size of previous columns in the same row = 5 * 4 = 20

Therefore, the address of A[4][5] would be 5000 + 1600 + 20 = 6620.

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8.

What will be the output of the program if the array begins at 65486 and each integer occupies 2 bytes? #include int main() { int arr[] = {12, 14, 15, 23, 45}; printf("%u, %u", arr+1, &arr+1); return 0; }

65488, 65490
64490, 65492
65488, 65496
64490, 65498

Correct Answer

A. 65488, 65496

Explanation

The program is using pointer arithmetic to print the memory addresses of the array elements.

In the printf statement, arr+1 is used to print the memory address of the second element in the array. Since each integer occupies 2 bytes, the memory address of the second element would be 65486 + 2 = 65488.

Similarly, &arr+1 is used to print the memory address of the element after the entire array. Since the array has 5 elements and each element occupies 2 bytes, the memory address of the element after the array would be 65486 + (5 * 2) = 65496.

Therefore, the output of the program would be 65488, 65496.

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9.

What will be the output of the program ? #include int main() { static int a[2][2] = {1, 2, 3, 4}; int i, j; static int p[] = {(int)a, (int)a+1, (int)a+2}; for(i=0; i<2; i++) { for(j=0; j<2; j++) { printf("%d, %d, %d, %d", ((p+i)+j), ((j+p)+i), ((i+p)+j), ((p+j)+i)); } } return 0; }

1, 1, 1, 12, 3, 2, 33, 2, 3, 24, 4, 4, 4
1, 2, 1, 22, 3, 2, 33, 4, 3, 44, 2, 4, 2
1, 1, 1, 12, 2, 2, 22, 2, 2, 23, 3, 3, 3
1, 2, 3, 42, 3, 4, 13, 4, 1, 24, 1, 2, 3

Correct Answer

A. 1, 1, 1, 12, 2, 2, 22, 2, 2, 23, 3, 3, 3

Explanation

The program declares a static 2D array "a" with dimensions 2x2 and initializes it with values 1, 2, 3, and 4. It also declares a static array of pointers "p" that points to the elements of "a".

In the nested for loop, the program prints the values of the elements in "a" using different combinations of pointer arithmetic.

The output shows the values of the elements in "a" in a specific order: 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3.

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10.

What will be the output of the program ? int 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); return 0; }

2, 1, 15
1, 2, 5
3, 2, 15
2, 3, 20

Correct Answer

A. 3, 2, 15

Explanation

The output of the program will be "3, 2, 15".

In the program, the array "a" is initialized with values {5, 1, 15, 20, 25}.

First, the value of a[1] is incremented by 1 using the prefix increment operator (++a[1]), so the value of a[1] becomes 2. This value is assigned to variable i.

Next, the value of a[1] is incremented by 1 again using the postfix increment operator (a[1]++), so the value of a[1] becomes 3. This value is assigned to variable j.

Then, the value of i (which is 2) is used as the index to access the element in array a, so m = a[2] which is 15.

Finally, the values of i, j, and m are printed as "3, 2, 15".

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Data Structure MCQ Exam Quiz!

What is the output of the code given below? #include void main() { char s= "hello"; char p = s; printf("%c\t%c", *(p + 3), s[1]); }

Quiz Preview

What will be the output of the program ? #include void fun(int **p); int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; int *ptr; ptr = &a[0][0]; fun(&ptr); return 0; } void fun(int p) { printf("%d", p); }

What does the following declaration mean? int (ptr)[10]*

What is the output of the code given below? #include void main() { float arr[5] = {12.5, 10.0, 13.5, 90.5, 0.5}; float ptr1 = &arr[0]; float ptr2 = ptr1 + 3; printf("%f ", *ptr2); printf("%d", ptr2 - ptr1);}

What is the output of the code given below? #include void main() { int arr[] = {10, 20, 30, 40, 50, 60}; int ptr1 = arr; int ptr2 = arr + 6; printf("Number of Elements : %d", (ptr2 - ptr1)); printf("Number of Bytes : %d", (char)ptr2 - (char) ptr1); }

What is the output of the code given below? # include void fun(int ptr) { ptr = 30; } int main() { int y = 20; fun(&y); printf("%d", y); }

Consider a two dimensional array int A[100][100]; // assume that an integer requires 4 bytes. Starting address of the array is 5000 What is the address of the element A[4][5] if the allocation is row major ?

What will be the output of the program if the array begins at 65486 and each integer occupies 2 bytes? #include int main() { int arr[] = {12, 14, 15, 23, 45}; printf("%u, %u", arr+1, &arr+1); return 0; }

What will be the output of the program ? #include int main() { static int a[2][2] = {1, 2, 3, 4}; int i, j; static int p[] = {(int)a, (int)a+1, (int)a+2}; for(i=0; i<2; i++) { for(j=0; j<2; j++) { printf("%d, %d, %d, %d", ((p+i)+j), ((j+p)+i), ((i+p)+j), ((p+j)+i)); } } return 0; }

What will be the output of the program ? int 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); return 0; }

Data Structure MCQ Exam Quiz!

What is the output of the code given below? #include void main() { char *s= "hello"; char *p = s; printf("%c\t%c", *(p + 3), s[1]); }

Quiz Preview

What will be the output of the program ? #include void fun(int **p); int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; int *ptr; ptr = &a[0][0]; fun(&ptr); return 0; } void fun(int **p) { printf("%d", **p); }

What does the following declaration mean? int (*ptr)[10]

What is the output of the code given below? #include void main() { float arr[5] = {12.5, 10.0, 13.5, 90.5, 0.5}; float *ptr1 = &arr[0]; float *ptr2 = ptr1 + 3; printf("%f ", *ptr2); printf("%d", ptr2 - ptr1);}

What is the output of the code given below? #include void main() { int arr[] = {10, 20, 30, 40, 50, 60}; int *ptr1 = arr; int *ptr2 = arr + 6; printf("Number of Elements : %d", (ptr2 - ptr1)); printf("Number of Bytes : %d", (char*)ptr2 - (char*) ptr1); }

What is the output of the code given below? # include void fun(int *ptr) { *ptr = 30; } int main() { int y = 20; fun(&y); printf("%d", y); }

Consider a two dimensional array int A[100][100]; // assume that an integer requires 4 bytes. Starting address of the array is 5000 What is the address of the element A[4][5] if the allocation is row major ?

What will be the output of the program if the array begins at 65486 and each integer occupies 2 bytes? #include int main() { int arr[] = {12, 14, 15, 23, 45}; printf("%u, %u", arr+1, &arr+1); return 0; }

What will be the output of the program ? #include int main() { static int a[2][2] = {1, 2, 3, 4}; int i, j; static int *p[] = {(int*)a, (int*)a+1, (int*)a+2}; for(i=0; i<2; i++) { for(j=0; j<2; j++) { printf("%d, %d, %d, %d", *(*(p+i)+j), *(*(j+p)+i), *(*(i+p)+j), *(*(p+j)+i)); } } return 0; }

What will be the output of the program ? int 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); return 0; }

What is the output of the code given below? #include void main() { char s= "hello"; char p = s; printf("%c\t%c", *(p + 3), s[1]); }

What will be the output of the program ? #include void fun(int **p); int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; int *ptr; ptr = &a[0][0]; fun(&ptr); return 0; } void fun(int p) { printf("%d", p); }

What does the following declaration mean? int (ptr)[10]*

What is the output of the code given below? #include void main() { float arr[5] = {12.5, 10.0, 13.5, 90.5, 0.5}; float ptr1 = &arr[0]; float ptr2 = ptr1 + 3; printf("%f ", *ptr2); printf("%d", ptr2 - ptr1);}

What is the output of the code given below? #include void main() { int arr[] = {10, 20, 30, 40, 50, 60}; int ptr1 = arr; int ptr2 = arr + 6; printf("Number of Elements : %d", (ptr2 - ptr1)); printf("Number of Bytes : %d", (char)ptr2 - (char) ptr1); }

What is the output of the code given below? # include void fun(int ptr) { ptr = 30; } int main() { int y = 20; fun(&y); printf("%d", y); }

What will be the output of the program ? #include int main() { static int a[2][2] = {1, 2, 3, 4}; int i, j; static int p[] = {(int)a, (int)a+1, (int)a+2}; for(i=0; i<2; i++) { for(j=0; j<2; j++) { printf("%d, %d, %d, %d", ((p+i)+j), ((j+p)+i), ((i+p)+j), ((p+j)+i)); } } return 0; }