Page 181 - C++

Address of an element (I,J) in row major = B+W(C(I-Lr)+(J-Lc ))

Therefore, 1500 = B+4(20(5-0)+(2- 0))

1500 = B+4(20*5+2)

1500 = B+4*102

1500 = B+408

B =1500-408

B=1092

Address of Arr[3][2] =1092+4(20*3+2)

=1092+4(62)

=1092+248

=1340.

(d) Write a function in C++ to perform a DELETE operation in a dynamically allocated queue considering the
following description: (4)
struct Node
{
float U, V ;
Node *Link ;
} ;
class QUEUE
{
Node *Rear, *Front ;
public :
QUEUE( )
{
Rear = NULL ;
Front = NULL ;
}
void INSERT( ) ;
void DELETE( ) ;
~ QUEUE( ) ;
} ;
Ans.

void Queue::DELETE( )

{

NODE *temp;

if(Front= = NULL)

cout<<”\nQueue Underflow”;