Question

6. Suppose a company decides to raise the salaries of its employees according to the following table:

employee_status years_of_service percent_raise
Full-time Less than 5 years 4.0
Full-time 5 years or more 5.0
Part-time Less than 5 years 2.5
Part-time 5 years or more 3.0
If employee_status value is ‘F’, the employee is Full-time; if it is ‘P’, he or she is a part-timer. Write this program that computes the new salary of an employee given his or her employee_status, year_of_service and salary by the users.
Input validation: Do not accept a negative value for the number of years_of_service,

Answer 1

help me to with this programming..its about selection =D

Answer 2

Do you know how to use if else and all?

Answer 3

yeah..i know...i hve do it..but cant get the answer =(


#include<stdio.h>
int main ()
{
int year;
char status;
float salary, new_salary;

printf ("Please type F (Full-Time) or P (Part-timer): ");
scanf ("%c", &status);
printf ("\nEnter your year of service: ");
scanf ("%d",&year);
printf ("\nPlease enter our current salary: RM");
scanf ("%.2f",&salary);

if (status =='F'&& year<5){
new_salary = 4.0 / 100 * salary + salary;
}
else if (status =='F'&& year>=5){
new_salary = 5.0 / 100 * salary + salary;
}
else if (status =='P'&& year<5)
{new_salary = 2.5 / 100 * salary + salary;
}
else if (status ='P'&& year>=5)
{new_salary = 3.0 / 100 * salary + salary;

}
printf ("Your new salary is : %.2f", new_salary);

return 0;
}

Answer 4

Ah i see your problem make that 5/100as 0.05 and 4/100 as 0.04 and so on.

Answer 5

This is because if the computer divides 5/100 it gives 0 as both numerator and denominator are integer type. If you make it 0.05 which is a float type it upgrades all other variables to the float type...

Answer 6

i have change it to 0.03 *salary + salary,
but still i cant get the answer =(

Answer 7

hmm i can't see any other problem with the code... maybe there's something in c i don't know c i know only c++

Answer 8

hurm...do c++ have large difference with c??

Answer 9

no but it is more easy to learn C++.I am learning C and whenever I get a problem I write the program in C++ and then convert the syntax into C.

Answer 10

I'll just try.
But by the way what selection are you talking about?

Answer 11

/*******************************************************************************

Suppose a company decides to raise the salaries of its employees according to the following table:

employee_status years_of_service percent_raise
Full-time Less than 5 years 4.0
Full-time 5 years or more 5.0
Part-time Less than 5 years 2.5
Part-time 5 years or more 3.0
If employee_status value is ‘F’, the employee is Full-time; if it is ‘P’, he or she is a part-timer. Write this program that computes the new salary of an employee given his or her employee_status, year_of_service and salary by the users.
Input validation: Do not accept a negative value for the number of years_of_service,

*************************************************************************/



#include<stdio.h>
#include<conio.h> /*For better view*/
#include<stdlib.h> /*For exit function*/
int main()
{
char time=0;
int yrs=0;
float sal=0,nsal=0;
printf("Welcome,do tou work full time(f) or part time(p): ");
scanf("%ch",&time);
printf("For how many years have you been working here?: ");
scanf("%d",&yrs);
printf("What you they pay you?(per month): RM");
scanf("%f",&sal);

/*Nested if for new salary*/

if(time=='f'&&yrs>5)
{
nsal=sal*0.05;
nsal+=sal;
}
else if(time=='f'&&yrs<=5)
{
nsal=sal*0.04;
nsal+=sal;
}
else if(time=='p'&&yrs>5)
{
nsal=sal*0.03;
nsal+=sal;
}
else if(time=='p'&&yrs<=5)
{
nsal=sal*0.025;
nsal+=sal;
}
else
{
printf("Something went wrong!!!Please check your output."); /*If input is no a number*/
exit(0);
}
printf("Your new salary is:RM %f",nsal); /*Change %f to %d if want an output in round figure*/
getch();
return 0;
}

Answer 12

thanks a lot..i'll learn C++ after C programming =D

Answer 13

We could advise you better if you showed us the answer that you *do* get...