C++ Primer Plus Chapter 7 Exercise 9

c plus plusChapter 7 ends with a short exercise on function pointers. The simplest way to complete this, which I provided in my source, is to create the two functions add() and calculate(), then call them in main() with a loop. The text mentions if you are feeling adventurous to include other functions in addition to add. For simplicities sake, I did the minimum requirement here. Finally,  we can put this chapter to rest. See my source below:

9. Design a function calculate() that takes two type double values and a pointer to a function that takes two double arguments and returns a double. The calculate() function should also be type double, and it should return the value that the pointed-to function calculates, using the double arguments to calculate(). For example, suppose you have this definition for the add() function: double add(double x, double y){return x + y}. Then, the function call in double q = calculate(2.5, 10.4, add); would cause calculate() to pass the values 2.5 and 10.4 to the add() function and then return the add() return value (12.9). Use these functions and at least one additional function in the add() mold in a program. The program should use a loop that allows the user to enter pairs of numbers. For each pair, use calculate() to invoke add() and at least one other function. If you are feeling adventurous, try creating an array of pointers to add()-style functions and use a loop to successively apply calculate() to a series of functions by using these pointers. Hint: Here’s how to declare such an array of three pointers: double (*pf[3])(double, double); You can initialize such an array by using the usual array initialization syntax and function names as addresses.

#include <iostream>

using namespace std;

double calculate(double x, double y, double (*pf)(double a, double b));
double add(double x, double y);

int main()
double x, y;
cout << "Enter two numbers (q to exit): ";
while(cin >> x >> y)
calculate(x, y, add);
cout << "The sum is: " << calculate(x, y, add) << "\n\n";
return 0;

double calculate(double x, double y, double (*pf)(double a, double b))
return pf(x, y);

double add(double x, double y)
return x + y;

C++ Primer Plus Chapter 7 Exercise 8

c plus plus 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')
student * ptr_stu = new student[class_size];
int entered = getinfo(ptr_stu, class_size);

for (int i = 0; i < entered; 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;
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";