Structured Query Language
ARR[15][20] is a two-dimensional array, which is stored in the memory along the row with each of its elements occupying 4 bytes. Find the address of the element ARR[5][15], if the element ARR[10][5] is stored at the memory location 35000.
ROW MAJOR:
Loc(ARR[I][J]) =BaseAddress + W [( I – LBR)*C + (J – LBC)]
(where W=size of each element = 4 bytes, R=Number of
Rows=15, C=Number of Columns=20 )
Assuming LBR = LBC = 0
LOC(ARR[10][5])
35000 = BaseAddress + W(I*C + J)
35000 = BaseAddress + 4(10*20 + 5)
35000 = BaseAddress + 4(205)
35000 = BaseAddress + 820
BaseAddress = 35000 - 820
= 34180
LOC(ARR[5][15])= BaseAddress + W(I*C + J)
= 34180 + 4(100 + 15)
= 34180 + 4 x 115
= 34180 + 460
= 34640
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ARR[15][20] is a two-dimensional array, which is stored in the memory along the row with each of its elements occupying 4 bytes. Find the address of the element ARR[5][15], if the element ARR[10][5] is stored at the memory location 35000.
Write function definition for WORD4CHAR() in C++ to read the content of a text file FUN.TXT, and display all those words, which has four characters in it.
Example:
If the content of the file fun.TXT is as follows:
When I was a small child, I used to play in the garden
with my grand mom. Those days were amazingly funful
and I remember all the moments of that time
The function WORD4CHAR() should display the following:
When used play with days were that time
Write a definition for function BUMPER( ) in C++ to read each object of a binary file GIFTS.DAT, find and display details of those gifts, which has remarks as “ÖN DISCOUNT”. Assume that the file GIFTS.DAT is created with the help of objects of class GIFTS, which is defined below:
class GIFTS
{
int ID;char Gift[20],Remarks[20]; float Price;
public:
void Takeonstock()
{
cin>>ID;gets(Gift);gets(Remarks);cin>>Price;
}
void See()
{
cout<<ID<<':'<<Gift<<':'<<Price<<':'<<Remarks<<endl;
}
char *GetRemarks(){return Remarks;}
}
Find the output of the following C++ code considering that the binary file MEM.DAT exists on the hard disk with a data of 1000 members.
class MEMBER
{
int Mcode;char MName[20];
public:
void Register();void Display();
};
void main()
{
fstream MFile;
MFile.open('MEM.DAT',ios::binary|ios::in);
MEMBER M;
MFile.read((char*)&M, sizeof(M));
cout<<'Rec:'<<MFile.tellg()/sizeof(M)<<endl;
MFile.read((char*)&M, sizeof(M));
MFile.read((char*)&M, sizeof(M));
cout<<'Rec:'<<MFile.tellg()/sizeof(M)<<endl;
MFile.close();
}
Give a suitable example of a table with sample data and illustrate Primary and Alternate Keys in it.
Consider the following tables CARDEN and CUSTOMER and answer (b) and (c) parts of question:
TABLE: CARDEN
Ccode |
CarName |
Make |
Color |
Capacity |
Charges |
501 |
A-Star |
Suzuki |
RED |
3 |
14 |
503 |
Indigo |
Tata |
SILVER |
3 |
12 |
502 |
Innova |
Toyota |
WHITE |
7 |
15 |
509 |
SX4 |
Suzuki |
SILVER |
4 |
14 |
TABLE: CUSTOMER
CCode |
Cname |
Ccode |
1001 |
Hemant Sahu |
501 |
1002 |
Raj Lal |
509 |
1002 |
Feroza Shah |
503 |
1004 |
Ketan Dhal |
502 |
Write SQL commands for the following statements:
Consider the following tables CARDEN and CUSTOMER and answer (b) and (c) parts of question:
TABLE: CARDEN
Ccode |
CarName |
Make |
Color |
Capacity |
Charges |
501 |
A-Star |
Suzuki |
RED |
3 |
14 |
503 |
Indigo |
Tata |
SILVER |
3 |
12 |
502 |
Innova |
Toyota |
WHITE |
7 |
15 |
509 |
SX4 |
Suzuki |
SILVER |
4 |
14 |
TABLE: CUSTOMER
CCode |
Cname |
Ccode |
1001 |
Hemant Sahu |
501 |
1002 |
Raj Lal |
509 |
1002 |
Feroza Shah |
503 |
1004 |
Ketan Dhal |
502 |
Give the output of the following SQL queries:
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