Program : 16

PROGRAM - 16

a)  Write a program to accept the element of the structure as: Employee-name and Basic pay.

Display the same structure along with the DA, CCA and Gross Salary for 5 employees.

Note: DA=51% of basic pay, CCA= Rs.100. Consolidated.

Program:

#include<stdio.h>

#include<conio.h>

void main()

{

     int i;

     struct emp

     {

        char name[20];

        int basic;

        int cca;

        int da;

        float gross;

     };

     struct emp e[5];

     clrscr();

     printf("\n enter emp name and basic:");

     for(i=0;i<5;i++)

     {

       scanf("%s %d",e[i].name,&e[i].basic);

       e[i].da=e[i].basic*0.51;

       e[i].cca=100;

       e[i].gross=e[i].basic+e[i].da+e[i].cca;

     }

     for(i=0;i<5;i++)

     {

       printf("\nName=%s\nBasics=%d\nCca=%d\nda=%d \nGross=%f",e[i].name,                          e[i].basic,e[i].cca,e[i].da,e[i].gross);

     }

     getch();

}

/*

INPUT:

 enter emp name and basic:

MICHAEL

1000

OUTPUT:

Name=MICHAEL

Basics=1000

Cca=100

da=510

Gross=1610.000000

*/

b)  Define a structure to store employee’s data with the following specifications: Employee-number, Employee-Name, Basic Pay, and Date of joining.

•   Write a function to store 10 employee details.

•   Write a function to implement the following rules while revising the basic pay.

i) If basic pay<= Rs. 5000 then increase it by 15%.

ii) If basic pay > Rs. 5000 and <=Rs. 25000 then      it    increase by 10%.

iii) If basic pay > Rs. 25000 then there is no change in basic pay.

  Write a function to print the details of employee who have completed 20 years of service from the date of joining.

Program:

#include<stdio.h>

#include<conio.h>

#include<math.h>

struct emp

{

     int eid;

     char ename[20];

     int basic;

     int day;

     int month;

     int year;

};

struct emp e[10];

int i;

void main()

{

     void empdetail();

     void basicpay();

     void display();

     clrscr();

     empdetail();

     basicpay();

     display();

     getch();

}

void empdetail()

{

     for(i=0;i<2;i++)

     {

        printf("\n enter id,name,basic,doj,enter day,month,year seperatly:");

        scanf("%d %s %d %d %d %d",&e[i].eid,&e[i].ename,&e[i].basic,&e[i].day,&e[i].month,&e[i].year);

     }

}

void basicpay()

{

     for(i=0;i<2;i++)

     {

        if(e[i].basic<=5000)

        e[i].basic=e[i].basic+((15*e[i].basic)/100);

        else if(e[i].basic>5000&&e[i].basic<=25000)

        e[i].basic=e[i].basic+((10*e[i].basic)/100);

        else

        e[i].basic=e[i].basic;

     }

}

void display()

{

     int year=2012;

     for(i=0;i<2;i++)

     {

         if((year-(e[i].year))>=20)

         printf("\nempid:%d\nname:%s\nsalary:%d\nday:%d\nmonth:%d\nyear:%d",

               e[i].eid,e[i].ename,e[i].basic,e[i].day,e[i].month,e[i].year);

     }

}

/*

INPUT:

enter id,name,basic,enter day,month,year seperatly of doj:

11       bittu       30000      13   04   1986

OUTPUT:

emp id:11

name:bittu

salary:30000

day:13

month:4

year:1986

*/