The operation for adding an entry to a stack is traditionally called:

Add
Append
Insert
Push

The main purpose of declaring a union.

To allow the same storage location to store data of different data types at different times.
To use it instead of structure.
To allow the same storage location to store data of different data types at the same time.
None of the above

The correct way for a structure in C to contain a pointer to itself is

typedef struct { char *item; NODEPTR next; } *NODEPTR;
struct node { char *item; struct node *next; }; typedef struct node *NODEPTR;
Both are correct.
None of above.

Consider the declaration :
struct sample
{
unsigned int a;
unsigned int b;
};
struct sample v;
The size of v.b is

1 bit
6 bits
5 bits
None of the above

Consider the program segment given below. (The italicized numbers represent program line numbers) :
1. #include
2. int y;
3. void main()
4. {
5. int x,*px,**ppx;void f1(int *);
6. x = 10;
7. y = 1000;
8. px = &x;
9. ppx = &px;
10. f1(px);
11. printf(“%d”,*px);
12. }
13. void f1(int *p)
14. {
15. *p = 20;
16. printf(“%d”,*p);
17. }
The printf() at line 11 prints the value:

10
1000
20
None of these options

#include
void prg(int);
main()
{
int a=10,j=3,i=2;
prg(a);
a*=( i + j );
printf(“%d”,a);
}
void prg(x)
{
int x;
int k=2;
x*=k;
return (x);
}
What is the output of this program?

10
20
50
None of these

This pointer for any class internally declared as

int *this;
X *thid;
const X *this;
X *const this;

What is the output of the following code?
#include
void main()
{
int arr[]={0,1,2,3,4,5,6};
int i,*ptr;
for(ptr=arr+4,i =0; i<=4; i++) printf("\n%d",ptr[-i]); }

Error
6 5 4 3 2
0 garbage garbage garbage garbage
4 3 2 1 0

#include
void main()
{
int I=10,*p=&I,**q=&p;
printf(“%u”,***(&(*(&Q))));
}
What is the output of this program?

value of p is printed
address of p is printed
value of I is printed
address of I is printed

How can I dynamically allocate a two-dimensional array?

int **array1 = (int **)malloc(nrows * sizeof(int *));
for(i = 0; i < nrows; i++)
array1[i] = (int *)malloc(ncolumns * sizeof(int));
int **array2 = (int **)malloc(nrows * sizeof(int *));
array2[0] = (int *)malloc(nrows * ncolumns * sizeof(int));
for(i = 1; i < nrows; i++)
array2[i] = array2[0] + i * ncolumns;
int *array3 = (int *)malloc(nrows * ncolumns * sizeof(int));
Any of the above.

Attempted

0

Correct

0

Wrong

0

Complete