Exercise 8 has us re-writing a skeleton program to fill in the missing functions. A combination of structs, pointers, and functions will complete this exercise. Comments describing where and what additions have been made are embedded in the code. See my solution below:
8. This exercise provides practice in writing functions dealing with arrays and structures. The following is a program skeleton. Complete it by providing the described functions:
#include <iostream>
using namespace std;
const int SLEN = 30;
struct student {
char fullname[SLEN];
char hobby[SLEN];
int ooplevel;
};
int getinfo(student pa[], int n);
void display1(student st);
void display2(const student * ps);
void display3(const student pa[], int n);
int main(){
cout << "Enter class size: ";
int class_size;
cin >> class_size;
while (cin.get() != '\n')
continue;
student * ptr_stu = new student[class_size];
int entered = getinfo(ptr_stu, class_size);
for (int i = 0; i < entered; i++) {
display1(ptr_stu[i]);
display2(&ptr_stu[i]);
}
display3(ptr_stu, entered);
delete [] ptr_stu;
cout << "Done\n";
return 0;
}
// getinfo() has two arguments: a pointer to the first element of
// an array of student structures and an int representing the
// number of elements of the array. The function solicits and
// stores data about students. It terminates input upon filling
// the array or upon encountering a blank line for the student
// name. The function returns the actual number of array elements
// filled.
int getinfo(student pa[], int n)
{
int i;
for(i = 0; i < n; i++)
{
cout << "Student Number: " << (i+1) << "\n";
cout << "Student's Name: ";
cin.getline(pa[i].fullname, SLEN);
cout << "Enter students hobby: ";
cin.getline(pa[i].hobby, SLEN);
cout << "Enter students OOP level: ";
cin >> pa[i].ooplevel;
cin.get();
}
return i;
}
// display1() takes a student structure as an argument
// and displays its contents
void display1(student st)
{
cout << "\n=============Display 1================\n";
cout << "Student name: " << st.fullname << "\n";
cout << "Hobby: " << st.hobby << "\n";
cout << "OOP Level: " << st.ooplevel << "\n";
cout << "=============End Display 1==============\n";
}
// display2() takes the address of student structure as an
// argument and displays the structure's contents
void display2(const student * ps)
{
cout << "\n============Display 2================\n";
cout << "Students Name: " << ps->fullname << "\n";
cout << "Hobby: " << ps->hobby << "\n";
cout << "OOP Level" << ps->ooplevel << "\n";
cout << "==========End Display 2================\n";
}
// display3() takes the address of the first element of an array
// of student structures and the number of array elements as
// arguments and displays the contents of the structures
void display3(const student pa[], int n)
{
for(int i = 0; i < n; i++)
{
cout << "\n============Display 3================\n";
cout << "Student: " << (i+1) << "\n";
cout << "Fullname: " << pa[i].fullname << "\n";
cout << "OOP Level: " << pa[i].ooplevel << "\n";
cout << "=============End Display 3============\n";
}
}
Demux 