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Data Structure
T[25][30] is a two dimensional array, which is stored in the memory along the row with each of its element occupying 2 bytes, find the address of the element T[10][15], if the element T[5][10] is stored at the memory location 25000.
LOC(T[10][15]) = LOC(T[5][10]) + 2(30*(10-5) + (15-10))
= 25000 + 2(150 + 5)
= 25000 + 2(155)
= 25000 + 310
= 25310
Write the definition of a member function ADDMEM() for a class QUEUE in C++, to add a MEMBER in a dynamically allocated Queue of Members considering the following code is already written as a part of the program.
struct Member
{
int MNO;
char MNAME[20];
Member *Next;
};
class QUEUE
{
Member *Rear,*Front;
public:
QUEUE(){Rear=NULL;Front=NULL;}
void ADDMEM();
void REMOVEMEM();
~QUEUE();
};
void QUEUE::ADDMEM()
{
Member *T;
T=new Member;
cin>>T->MNO;
gets(T->MNAME);
T->Next=NULL;
if (Rear==NULL)
{
Rear=T;Front=T;
}
else
{
Rear->Next=T;
Rear=T;
}
}
T[20][50] is a two dimensional array, which is stored in the memory along the row with each of its element occupying 4 bytes, find the address of the element T[15][5], if the element T[10][8] is stored at the memory location 52000.
Loc(T[I][J])
=BaseAddress + W [( I – LBR)*C + (J – LBC)]
(where
W=size of each element = 4 bytes,
R=Number of Rows=20, C=Number of Columns=50)
Assuming LBR = LBC = 0
LOC(T[10][8])
52000 = BaseAddress + W[ I*C + J]
52000 = BaseAddress + 4[10*50 + 8]
52000 = BaseAddress + 4[500 + 8]
52000 = BaseAddress + 4 x 508
BaseAddress = 52000 2032
= 49968
LOC(T[15][5])= BaseAddress + W[ I*C + J]
= 49968 + 4[15*50 + 5]
= 49968 + 4[750 + 5]
= 49968 + 4 x 755
= 49968 + 3020
= 5298
Write the definition of a member function INSERT() for a class QUEUE in C++, to insert an ITEM in a dynamically allocated Queue of items considering the following code is already written as a part of the program.
struct ITEM
{
int INO; char INAME[20];
ITEM *Link;
};
class QUEUE
{
ITEM *R,*F;
public :
QUEUE() {R=NULL;F=NULL;}
void INSERT();
void DELETE();
~QUEUE();
};
void QUEUE::INSERT()
{
ITEM *T = new ITEM;
cin>>T->INO;
gets(T->INAME);
T->Link = NULL;
if(R==NULL)
{
F=T; R=T;
}
else
{
R->Link=T; R=T;
}
}
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