Question

4. What is stored in each frame? Circle all the correct answers

a. Local variables

b. Static variables

c. Return address

d.Function parameters

e. Global variables

f. Function code

5. A   union   is   a   data   type   that         
  
A. can   only   be   declared       inside   a   struct   type  

B. is   useful   for   constructing   linked   lists  

C. can   only   be   used   with   arrays  

D. reserves   the   same   space   of   memory   for   its       fields  

E. None   of   the   above  

6. What will be the output of the following code Short age [6] = {25, 28, 29, 30, 31,32};
int *p = NULL;

p = &age[4];
p += 2;

printf(“%d\n”, *p);

A. 28 B. 29 C. 30 D. 31 E. None of the above

7. Which of the following statements is an array of 10 pointers to an int:

A. int arr[10]*;

B. int *arr[10];

C. int &arr[10];

D. int arr[10]&;

E. int arr[10];

F. None of the above

8. Given the following code, which statement will print True
int x = 5;

int y = 6;
int *p = &x;

int *q = &y int **pp = &p;

int **qq = &q;

A. If (*pp == p) printf (“True\n);

B. If (**pp == *q -1) printf (“True\n”);

C. If (*q == *p) printf (“True\n”);

D. If (*pp == *p) printf (“True\n”);

E. If (*pp == x) printf (“True\n”);
F. If (**qq == y) printf (“True\n”);

G. None of the above

Answer 1

4.

Each frame contain:

• Local variables
• Static variables
• Function parameters
• Global variables

5.

• A union reserves the same space of memory for its fields.
• It is used in the c language.
• It is also appeared like as the structures but difference is reserves the same space of memory for its fields.

Option D

6.

int age [6] = {25, 28, 29, 30, 31,32};

int *p = NULL; // Initially pointer is NULL

p = &age[4]; // it contains the value 31

p+=2; // Pointer goes to next position by default its value is 0

printf("%d\n", *p); // It print 0

Option E

7.

int *arr[10]; contains the array of 10 pointers.

• All pointers may be type of integer.

Option B

8.

int x = 5;

int y = 6;
int *p = &x;

int *q = &y; int **pp = &p;

int **qq = &q;

Options:

If (**pp == *q -1) printf (“True\n”);

If (**pp == *q -1) printf (“True\n”);

If (**qq == y) printf (“True\n”);

Options A, B and F are correct.