PL/SQL PROGRAM TO FIND BIG NUMBER

declare
a number:=&a;
b number:=&b;
c number:=&c;
begin
if( ( a > b) and ( a > c )) then
dbms_output.put_line('a is big');
elsif(( b > a ) and ( b > c )) then
dbms_output.put_line('b is big');
elsif(( c > a ) and ( c > b )) then
dbms_output.put_line('c is big');
end if;
end;




OUTPUT:


Enter value for a: 12
old   2: a number:=&a;
new   2: a number:=12;
Enter value for b: 50
old   3: b number:=&b;
new   3: b number:=50;
Enter value for c: 90
old   4: c number:=&c;
new   4: c number:=90;
c is big

MATRIX ADDITION USING COBOL LANGUAGE

       IDENTIFICATION DIVISION.
       PROGRAM-ID MAT-ADD.
       ENVIRONMENT DIVISION.
       DATA DIVISION.
       WORKING-STORAGE SECTION.
        01 MAT1.
           02 MR1 OCCURS 5 TIMES.
            03 MC1 PIC 9(2) OCCURS 5 TIMES.
        01 MAT2.
           02 MR2 OCCURS 5 TIMES.
            03 MC2 PIC 9(2) OCCURS 5 TIMES.
        01 MAT3.
           02 MR3 OCCURS 5 TIMES.
            03 MC3 PIC ZZZ(9) OCCURS 5 TIMES.
           77 ROW1 PIC 9.
           77 ROW2 PIC 9.
           77 COL1 PIC 9.
           77 COL2 PIC 9.
           77 I PIC 9.
           77 J PIC 9.
       PROCEDURE DIVISION.
       MAIN-PARA.
           DISPLAY 'MATRIX ADDITION'.
           DISPLAY '  '.
           DISPLAY 'ENTER THE NUMBER OF ROW AND COLUMNS OF 1 MATRIX'.
           ACCEPT ROW1.
           ACCEPT COL1.
           DISPLAY 'ENTER THE NUMBER OF ROW AND COLUMNS OF 2ND MATRIX'.
           ACCEPT ROW2.
           ACCEPT COL2.
           IF(ROW IS NOT EQUALS TO ROW2) OR (COL1 NOT EQUALS TO COL2)
           DISPLAY 'MATRIX ADDITION IS NOT POSSIBLE'
           ELSE
           PERFORM COMPU-PARA
           STOP RUN.
       COMPU-PARA.
           DISPLAY 'ENTER THE ELEMENTS OF MATRIX'.
           PERFORM ACCPARA1 VARYING I FROM 1 BY 1 UNTIL I > ROW1
           AFTER J FROM 1 BY 1 UNTIL J > COL1.
           DISPLAY 'ENTER THE ELEMENTS OF MATRIX2'.
           PERFORM ACCPARA2 VARYING I FROM 1 BY 1 UNTIL I > ROW2
           AFTER J FROM 1 BY 1 UNTIL J > COL2.
           PERFORM ADDMAT-PARA VARYING I FROM 1 BY 1 UNTIL I > ROW1
           AFTER J FROM 1 BY 1 UNTIL J > COL1.
           DISPLAY ( 1 1 ) ERASE.
           MOVE 3 TO LIN.
           MOVE 10 TO COL.
           DISPALY ' '.
           DISPLAY ' THE FIRST MATRIX IS'.
           DISPLAY ' '.
           PERFORM DISP1-PARA VARYING  I FROM 1 BY 1 UNTIL I > ROW1
           AFTER J FROM 1 BY 1UNTIL J > COL1.
           DISPLAY ' '.
           MOVE 10 TO LIN.
           MOVE 10 TO COL.
           DISPLAY ' '.
           DISPLAY 'THE SECOND MATRIX IS'.
           DISPLAY ' '.
           PERFORM DISP2-PARA VARYING  I FROM 1 BY 1 UNTIL I > ROW12
           AFTER J FROM 1 BY 1UNTIL J > COL1.
           DISPLAY ' '.
           MOVE  17 TO LIN.
           MOVE  10 TO COL.
           DISPLAY ' '.
           DISPLAY 'THE RESULTANT MATRIX IS'.
           DISPLAY '  '.
           PERFORM DISP-PARA VARYING  I FROM 1 BY 1 UNTIL I > ROW 1
           AFTER J FROM 1 BY 1UNTIL J > COL1.
           DISPLAY ' '.
       ACCPARA1.
           ACCEPT MC1 (I J).
       ACCPARA2.
           ACCEPT MC2 (I J).
       ADDMAT-PARA
           COMPUTE MC3 (I J) = MC1 (J I) + MC2 (I J).
       DISP-PARA
           DISPLAY (LIN . COL ) MC3 (I J).
           ADD 5 TO COL.
           IF J IS EQUAL TO COL
           ADD 1 TO LIN MOVE 10 TO COL.
       DISP1-PARA
           DISPLAY (LIN , COL ) MC1 (I J).
           ADD 5 TO COL.
           IF J IS EQUAL TO COL.
           ADD 1 TO LIN MOVE 10 TO COL
       DISP2-PARA.
           DISPLAY ( LIN, COL) MC1 (I J).
           ADD 5 TO COL.
           IF J IS EQUAL TO COL.
           ADD 1 TO LIN MOVE 10 TO COL
         
output:
  
enter the number of rows for 1 matrix:
2*2
enter the number of rows for 2 matrix:
2*2
enter the elements of 1 matrix:
1 2
3 4
the first matrix is
1 2
3  4
enter the second matrix:
3 4
6 7
the second matrix is:
3 4
6  7
the resultant matrix is:
4 6
9 11

FIND OUT THE BIG VALUE USING JAVA PROGRAM

import java.io.*;
import java.lang.*;
class Bigvalue
{
public static void main(String[]args)
{
int a=5;
int b=2;
int c=7;

if((a>b)&&(a>c))
System.out.println("A is big");
else
if((b>a)&&(b>c))
System.out.println("b is big");
else
if((c>b)&&(c>a))
System.out.println("C is big");
}
}

SIMPLE COBOL PROGRAM TO FIND OUT THE STUDENT TOTAL AND GRADES

     
/*COBOL  PROGRAM TO FIND OUT STUDENT GRADES. */

       IDENTIFICATION DIVISION.
       PROGRAM-ID STUDENTGRADE.
       ENVIRONMENT DIVISION.
       DATA DIVISION.
       WORKING-STORAGE SECTION.
           77 M1 PIC 9(2).
           77 M2 PIC 9(2).
           77 M3 PIC 9(2).
           77 TOTAL PIC 9(4).
       88 FIRST1 VALUE 180 THRU 300.
       88 SECOND VALUE 150 THRU 179.
       88 THIRD VALUE 120 THRU 149.
       88 FAIL VALUE 0 THRU 119.
       PROCEDURE DIVISION.
       PARA.
           DISPLAY 'ENTER STUDENTMARKS'.
           ACCEPT M1.
           ACCEPT M2.
           ACCEPT M3.
           COMPUTE TOTAL = M1 + M2 + M3.
           DISPLAY TOTAL.
           IF FIRST1
           DISPLAY 'A GRADE'
           ELSE
           IF SECOND
           DISPLAY 'B GRADE'
           ELSE
           IF THIRD
           DISPLAY 'C GRADE'
           ELSE
           DISPLAY 'FAIL'
           STOP RUN.  



OUTPUT:

ENTER STUDENT MARKS

45

55

65

0165

B GRADE

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SIMPLE JAVA PROGRAM FO ADDITION OF TWO NUMBERS USING 3 VARIABLES

SIMPLE JAVA PROGRAM FO ADDITION OF TWO NUMBERS
 USING 3 VARIABLES

import java.lang.*;
class add
{
public static void main(String args[])
{
int a=10,b=20,temp;// HERE A,B,TEMP IS VARIABLES
temp=a+b;//TEMP VARIABLE STORES ADDITION OF A,B
System.out.println("temp="+temp);
}
}



OUT PUT:-
temp=30

Difference between compiler and interpreter

Compiler

A compiler is a piece of code that translates the high level language into machine language. When a user writes a code in a high level language such as Java and wants it to execute, a specific compiler which is designed for Java is used before it will be executed. The compiler scans the entire program first and then translates it into machine code which will be executed by the computer processor and the corresponding tasks will be performed.  

Shown in the figure is basic outline of the compilation process, here program written in higher level language is known as source program and the converted one is called object program. 

Interpreter

Interpreters are not much different than compilers. They also convert the high level language into machine readable binary equivalents. Each time when an interpreter gets a high level language code to be executed, it converts the code into an intermediate code before converting it into the machine code. Each part of the code is interpreted and then execute separately in a sequence and an error is found in a part of the code it will stop the interpretation of the code without translating the next set of the codes.  

The differences between compiler and interpreter are :-