Employee
is a type name and itself doesn’t have members to read data in.
Use the variable to access members.
Additionally, you may want to use a reference to modify caller’s local variable.
Add &
after the type name to use reference.
void Setter(Employee& E) //function for setting value in Employees
{
cout<<"Enter Id:";
cin>>E.Id;
cout<<"Enter Name:";
cin>>E.Name;
cout<<"Enter Gender:";
cin>>E.Gender;
cout<<"Enter Designation:";
cin>>E.Des;
cout<<"Enter Date of joining(DD/MM/YYYY):";
//cin>>E.Date.day>>E.Date.month>>E.Date.year;
}
Note that the comment-outed line
cin>>E.Date.day>>E.Date.month>>E.Date.year;
is wrong because the type of E.Date
is int
and it won’t have members. You will have to alter the strucure’s declaration to let the structure hold a date (or three additional integers).