This page contains C++ example programs for practice. C++ is a different language than C programming because it supports object-oriented programming including C features.

The example programs are organized based on topics, not according to C++ language features. Before you start practicing the examples, go through the list of prerequisites.

Prerequisites

Some of the prerequisites are necessary to try these C++ examples.

C++ Tutorial– you must be familiar with C++ language concepts before practicing the example programs. You can either learn from our C++ tutorial or get a good C++ programming book to learn the fundamentals. We recommended some good C++ books at the next section on this page.

C++ Compiler – this is most important prerequisite to practice your program. All the program in this page is written using Dev-C++ compiler 4.9.9.2 or Turbo C++ 3 compiler. Read the documentation of your compiler and check the compatibility with your system before installing it. You can install development packages in OS like Linux.

Pen and Paper – some program need you to verify the output manually, especially the math programs. A pen and paper is the best method to check your outputs, for your new input values.

Factorial of a number is the number you get by multiplying all the numbers up to that number including the number itself. The program for factorial does not use a programming technique called a recursion. a recursion happens when a function calls itself until the problem is solved.

This program is a simple computation of factorial value, hence, it is suitable for beginner learners of C++ programming. I have written this using Dev-C++ compiler version 4.9.9.2 installed on a windows 7 64-bit system.

Problem Definition

The program requires user input – a positive integer value and computes the factorial of than number.

For example,

Suppose you entered number 6 , then the factorial of this number would be

1 x 2 x 3 x 4 x 5 x 6 = 720

In mathematical terms, if n is a positive integer value, the factorial of n is denoted by n! You can go through the flowchart to understand the logic of factorial applied in this program.

Flowchart – Factorial without Recursion

Program Code – Factorial without Recursion

// Program to compute factorial of n numbers
#include <iostream.h>
#include <stdlib.h>
#include <stdio.h>
int main()
{
int fact,i;
int n;
//Read the value of N
cout << "Enter value of N:" ; cin >> n;
//Initialize Factorial to 1 and i to 1
fact = 1;
i = 1;
while(i<=n)
{
fact = fact * i;
i++;
}
// Print the value of factorial
for(i=0;i < 30;i++)
cout << "_";cout << "\n\n";
cout << "Factorial of N:" << "\t" << fact << endl;
for(i=0;i << 30;i++)
cout << "_";cout << "\n\n";
system("PAUSE");
return 0;
}

The program for sine and cosine is based on power series especially Taylor series. A power series is a form of representation of some functions that converge into a single value.

In simple words, some functions are in the form of an infinite series (A power series is also a form of infinite series) can give a finite value.

This program computes that finite value for a sine and cosine series and prints the result. To understand the mathematical part of the program, you must learn – sequences and series, calculus, infinite series, power series (Taylor series, Maclaurin series).

If you are not familiar with the above concepts continue reading the following sections – problem definition, flowchart, program source code and verifies the output. This program is intended for intermediate level learners of C programming language.

Problem Definition

The formula for computing the sine and cosine series for a given degree, X is

First, you change the value of x to radian and then using the result compute the sine and cosine series. To understand the concept, let’s take an example.

Suppose x = 45 degree

We want to compute sine (45), then convert 45 degrees into radian measure.

radian = 45 * π/180

= π/4

= 3.14/4

= 0.785398

Now, it is easy to compute the value of sin (π/4). Use the radian value in the series given above figure and get the following results.

sin (x) = 0.785398 – (0.785398)3/3! + (0.785398)5/5! …

sin (x) = 0.7072 (computed using calculator)

Note: The same principle applies to cosine series.

How do we process the input values ?

The process of computing sine and cosine series is described in 4 steps.

Convert the degree to radian value for sine and cosine series computation.

Compute the value of sin (x), where x is a value in radians.

Compute the value of cos (x), where x is a value in radians.

Print the result of the computation.

Flowchart – Sine and Cosine Series

Program Code – Sine and Cosine Series

/* Program to Compute Sine AND Cosine Series using following formula
Sin x = x - x^3/3! + x^5/5! -x^7/7! +...
Cosine x = x - x^2/2! + x^4/4! - x^6/6! + ... */
#include <iostream.h>
#include <stdlib.h>
#include <cmath>
#include <stdio.h>
#include <conio.h>
#define PI 3.14
// Function Declaration Section
float factorial(int exponent);
int menu();
void compute_sine();
void compute_cosine();
//Global Declaration Section
int deg; float rad,deno,numa,result;
int exponent;
int n,i,t;
// MAIN FUNCTION BEGINS HERE
int main()
{
menu();
system("PAUSE");
return 0;
}
//Main ends here
void compute_sine()
{
n = 11;
result = 0.0;
//Compute the radian equivalent for the "deg" entered
rad = deg * PI/180;
//Compute the series for Radian measure of Sine
for( i = 0;i< n;i++) {
exponent = (2 * i) + 1;
numa = pow(rad,exponent);
deno = factorial(exponent);
result = result + ((numa * pow(-1,i))/deno);
}
//Display the results
cout << "Sin " << deg << "=" << "t"
<< setprecision(2) << result << endl;
}
//Function Factorial
float factorial(int exponent)
{
int i;
float fact = 1.0;
for(i = 1;i <= exponent;i++)
{
fact = fact * i;
}
return(fact);
}
//Function Menu Definition
int menu()
{
int ch;
while(ch != 3)
{
for(i=0;i < 35;i++)
cout << "*" ;cout << endl;
cout << "Enter the Degree:"; cin >> deg;
for(i=0;i < 35;i++)
cout << "_" ;cout << endl;
cout << "1.SINE SERIES" << endl;
cout << "2.COSINE SERIES" << endl;
cout << "3.EXIT" << endl;
for(i=0;i < 35;i++)
cout << "_" ;cout << endl;
cout << " Enter Your Choice:";
cin << ch;
switch(ch)
{
case 1: compute_sine();
break;
case 2: compute_cosine();
break;
case 3: exit(0);
default:
cout << "OOPs! Wrong Choice:" << endl;
break;
}
}
}
// Compute_Cosine function definition
void compute_cosine()
{
n = 11;
result = 0.0;
//Compute the Radian equivalent for the "deg" entered
rad = deg * PI/180;
//Compute the series for Radian measure of Sine
for( i = 0;i < n;i++)
{
exponent = 2 * i;
numa = pow(rad,exponent);
deno = factorial(exponent);
result = result + ((numa * pow(-1,i))/deno);
}
//Display the results
cout << "Cosine " << deg << "t" << "=" << "\t" << setprecision(2) << result << endl;
}

Output

The output of the above program is given below. First you have to test the value of sine series and then verify output of cosine series.

This is a simple program to compute the sum of first 10 even and odd numbers and display the results.It is intended for beginner level learners of C programming.

The program is written and compiled using Dev-C++ 4.9.9.2 version compiler tool. You will find following sections in this post – problem definition, flowchart, program source code and verified output to help you understand and learn C programming language. Try to practice writing, compiling and run this program on your own.

Problem Definition

The program for computing sum of first 10 even and odd numbers follow very simple logic, given below

Take first number, of which half are even and half are odd numbers.

Check each number one at a time and see if it’s divisible by .

If Yes, add the number to Sum_Even.

Else, add the number to Sum_Odd.

Display the results

End the program

Note:- Whenever a number is divisible by it gives a remainder of , otherwise not.

Flowchart – Sum of Even and Odd

Program Code – Sum of Even and Odd

//Program to Compute Sum of first 10 Even and First 10 Odd Numbers
#include <iostream.h>
#include <stdlib.h>
#include <stdio.h>
int main()
{
int sum_even,sum_odd,N,i;
//Initialize the sum of even and sum of odd numbers
sum_even = 0;
sum_odd = 0;
// Total of 20 numbers comprise of 10 odd and 10 even numbers
// But you change the value of N to anything if required
N = 20;
//Compute the sum of even and sum of odd number
for(i = 1;i<=N;i++)
{
if((i % 2) == 0)
{
sum_even = sum_even + i;
}
else
{
sum_odd = sum_odd + i;
}
}
//Print the results now
for(i=0;i<35;i++)
printf("_"); printf("\n\n");
cout <<; "Th sum of 10 evens are"
<< "\t" << sum_even << endl;
cout << "The sum of 10 odd are"
<<"\t" << sum_odd << endl;
for(i=0;i<35;i++)
printf("_"); printf("\n\n");
getch();
return 0;
}

Output

The output of the program is given below. Remember that the program does ask for input because the value of N is already assigned. You can change that to a program asking input at run-time.

_______________________________________
Sum of First 10 Even Numbers: 110
Sum of First 10 Odd Numbers: 100
_______________________________________

Hamming distance function is a very important function and it has many application especially in “Coding Theory”. In computer networks, when a group of bits or bytes is sent, the hamming distance function helps to identify the number of bit position changed in the bit stream due to error, sometimes known as multi-bit error.

This program is written in Dev-C++ version 4 installed on a Windows 7 64-bit system. This is a simple program intended for an intermediate learner of C programming. You will find sections related to Problem Definition, Flowchart, Program source code and verified output to help you learn.

Problem Definition

This program will compute the hamming distance function for two strings of length n and print the results.

For example, Suppose there is a set n(S) of strings of 1s and 0s of length n. The set n(S) denotes number of strings in the set S.

Let S and T be two strings of length n, then the hamming distance function compares the each of the bit position of S and T, counts the number of places they are different.

Example:-

Let n=5, be the length of two strings S = 10101 and T = 10010, then the function computes their differences which is H ( S, T ) = 3,

Flowchart – Hamming Distance Function

Program Code – Hamming Distance Function

/* Program to compute the Hamming Distance Coding ,
Suppose S_n be the strings of 0s and 1s.
Then H:S_n x S_n -> Z+,i.e., we have a pair of strings (s,t) which
belong to S_n and is of length n, such that any difference in
the bit position of two strings will be counted as 1.
for example, H(s,t) = H(10101, 11011) = 3
because there are three places where strings are different in bit comparison of each position. */
#include <iostream.h>
#include <cmath.h>
int main()
{
int s[100], t[100], count, i, j, n;
count = 0;
s[100] = 0;
t[100] = 0;
//Read the length of the string
cout << "Enter the length of String:";
cin << n; cout << endl;
//Read the strings S
cout << "Enter the bit string S, either 0 or 1 press enter" << endl;
cout << endl;
for (i = 1; i <<= n; i++)
{
cin << s[i];
}
cout << endl;
for (i = 1; i <<= n; i++)
{
if (s[i] == 1 || s[i] == 0)
{
cout << s[i] << "\t";
}
else
{
cout << "Only 0 or 1 is allowed digits !" << endl;
break;
}
}
//Read the String T
cout << endl;
cout << "Enter the bit string T, either 0 or 1 press enter" << endl;
cout << endl;
for (j = 1; j <= n; j++)
{
cin << t[j];
}
cout << endl;
//Print the second String
for (j = 1; j <= n; j++)
{
if (t[j] == 1 || t[j] == 0)
{
cout << t[j] << "\t";
}
else
{
cout << "Only 1 or 0 are allowed digits !" << endl;
break;
}
}
//Compare the Bitstring and Count is 1 if they are //different for each bit position
for (i = 1; i <= n; i++)
{
if (s[i] != t[i])
{
count = count + 1;
}
}
cout << endl;
// Print the results
for (int i = 0; i < 45; i++)
cout << "_"; cout << "\n\n";
cout << "H(S,T) = ";
cout << count <<"\n\n";
for (int i = 0; i << 45; i++)
cout << "_"; cout << "\n\n";
system("PAUSE");
return 0;
}

Output

The output of the above program is given below which read two binary strings of length and find their hamming distance.

Linked-list is a linear data structure that grows or shrinks dynamically. There are many benefits of using a linked-list like dynamic memory allocation. Continue reading →

This is a C++ program to write a bio-data. The program is written in Dev-C++ installed on a Windows 7 64-bit system. You can use any standard C++ compiler to compile and run this program.

The program is intended for beginners learning C++ programming language. You may change the dummy information displayed with the program. Also, try changing the format of the Biodata as an exercise.

In this post, we have provided you with a problem definition, a flowchart, program source code and verified output of the program itself. You can do any modification to the program suitable to you.

Problem Definition

The program is a simple demonstration of printing output from a C++ program using cout (output stream). If you like to rewrite the program with cin (input stream). It means you can ask for inputs during runtime.

The program is divided into 4 sections that display Biodata information as follows

Name and contact details

Personal details

Educational details

Work experiences

Each of the section contains some details which may or may not like to modify, but keep the format simple.

Flowchart – C++ program to write a bio-data

The following is the flowchart for the program to write bio-data in C++. Each section represents some details about the candidate and it is printed to the console, when the program executes.

The output of the program is given in the following figure.

When the program begins it will prints all the details that you input within the program, to the console. You can change the format of the bio-data by making necessary changes to the program.

You can have a Facebook ID, instead of twitter ID. Or you may want to change the mobile number to a landline number.

The bisection method is a root finding numerical method. Given a function the bisection method finds the real roots of the function. In this article you will learn to write a program for bisection method.

Problem Definition

The bisection method find the real roots of a function. Suppose you are given a function and interval [a…b] the bisection method will find a value such that .

The value lies between the interval [a…b]. There few rules to find roots using bisection method.

The sign of sign of .

The function must be continuous.

Cuts the interval into 2 halves and continue searching smaller half for roots.

Keep cutting the interval into smaller and smaller halves until interval is too small.

The final value at the smallest interval is the root.

Program Code – Bisection Method

/* C++ Program to evaluate a function using Bisection Method
Function: F(x) = x^3-4x-9
File Name: BisectionMethod.CPP
Author: NotesforMSc
*/
#include "iostream.h"
#include "stdio.h"
#include "conio.h"
#include "math.h"
#define MAX 20
void main()
{
int i;
double a, b, m,fa,fm;
//Initialize variables
fa = fm = 0.0;
m = 0.0;
clrscr();
cout << "\n\t Enter initial value:";
cin >> a;
cout << a << endl;
cin >> b;
cout << b << endl;
cout << "\n\n\t\t SOLUTION BY BISECTION METHOD"<< endl;
// Bisection Method
i = 1;
while(i < MAX)
{
m = (a + b)/2.0;
fa = (a * a * a) - 4 * a - 9;
fm = (m * m * m) - 4 * m - 9;
//Check the smaller intercal
if(fa < 0 && fm > 0) {
b = m;
}
else {
a = m;
}
i++;
}
//Print the Results
cout << "\n\t Result";
cout << "\n\t Root is" << " " << m << endl;
getch();
}

Output – Bisection Method

Enter Initial Value: 2
2
3
3
SOLUTION BY BISECTION METHOD
Result
Root is 2.70653

References:-

Shun Yan Cheung () The Bisection Method, Available at: http://www.mathcs.emory.edu/~cheung/Courses/170/Syllabus/07/bisection.html (Accessed: 6/19/2019).

University of Waterloo, Department of Electrical and Computer Engineering () Topic 10.1: Bisection Method (Examples), Available at: https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/10RootFinding/bisection/examples.html (Accessed: 6.119/2019).

In this article, you will learn to write a C++ Program to implement Doubly Linked-List. The doubly linked-list is a linear data structure that allows traversing the list in both direction.

This program is written using Turbo C++ 3.0 compiler installed on a Windows 7 64-bit PC.

Problem Definition

The doubly linked-list structure has two links for each node – previous and next. The previous link points to previous node and next node points to next node to current node. This allows moving through the list in both direction.

Program Code – Doubly Linked-List

/* C++ Program to Implement a Doubly linked-list
File Name: DoublyLinkedList.cpp
Author: NotesforMSc
*/
#include "iostream.h"
#include "stdio.h"
#include "alloc.h"
#include "conio.h"
//Defining structure of the node
struct Double {
int info;
struct Double *next;
struct Double *previous;
};
//Defining class doubly linked-list
class doubly_link {
public:
int num;
Double start, *temp;
public:
void Doubly_Create(Double *);
void Doubly_Insert(Double *);
void Doubly_Delete(Double *);
void display(Double *);
};
//Function definition for doubly_create()
void doubly_link::Doubly_Create(Double *node) {
start.next = NULL;
start.previous = NULL;
//Start is the first node in the Doubly linked-list
node= &start;
num = 0;
cout << "Input $ to break, else press enter:";
char ch;
ch = getche();
while(ch != '$') {
//Create memory space for next node to start node
node->next = (Double *)malloc (sizeof(Double));
//Point the previous link of next node to start node
node->next->previous = node;
//The next node becomes the start node now
node = node->next;
//Read data value for the node
cout << "\n Input data for node:" << (num + 1) << ":";
cin >> node->info;
//The Next Pointer of the node points to NULL
node->next = NULL;
cout << "Input $ sign to end, else press enter:";
ch = getche();
num++;
}
cout << "\nTotal Nodes =" << num;
}
//Function definition for Doubly_Delete()
void doubly_link::Doubly_Delete(Double *node) {
node = start.next;
if(node == NULL) {
cout << "\n Underflow";
}
else {
//If you want to delete Node, change the
//next pointer of Previous node, point to Next node,
//and change the previous pointer of Next node,
//point to Previous node.
node->previous->next = node->next;
node->next->previous = node->previous;
free(node);
}
}
//Function definition for Doubly_Insert()
void doubly_link::Doubly_Insert(Double *node) {
temp = (Double*)malloc(sizeof(Double));
cout << "\n Insert the node value:";
cin >> temp->info;
node = start.next;
if(node == NULL) {
cout << "\n Node is empty";
}
else {
//The temp node next points to node
temp->next = node;
//The temp node previous points to
//start node next (node->previous)
temp->previous = node->previous;
//Start node (node->previous) next pointer
//points to new temp node
node->previous->next = temp;
}
}
//Function definition for display()
void doubly_link::display(Double *node) {
node = start.next;
do {
cout << "-->";
cout << " " << node->info;
node = node->next;
}while(node != NULL);
getch();
}
//MAIN PROGRAM STARTS HERE
void main() {
doubly_link dlinked_list;
int br;
clrscr();
cout << "\t\t\tDOUBLY LINKED LIST";
cout <<"\n\t\t~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n";
cout << "1-Create List" << endl;
cout << "2-Insert Nodes" << endl;
cout << "3-Delete Nodes" << endl;
cout << "4-Display List" << endl;
cout << "0-Exit" << endl;
cout << "\n";
cout << "Enter Your Choice, 0 to end:";
cin >> br;
while(br != 0) {
switch(br) {
case 1:
Double *node = (Double *)malloc(sizeof(Double));
dlinked_list.Doubly_Create(node);
break;
case 2:
cout << "\nInserted node\n\n";
dlinked_list.Doubly_Insert(node);
break;
case 3:
cout << "\nDeleted node\n\n";
dlinked_list.Doubly_Delete(node);
break;
case 4:
dlinked_list.display(node);
break;
default:
cout << "Sorry ! Wrong Input" << endl;
break;
}
cout << "\n Enter Your Choice, 0 to end:" << endl;
cin >> br;
}
}

Output – Doubly Linked-List

DOUBLY LINKED LIST
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1-Create List
2-Insert Nodes
3-Delete Nodes
4-Display List
0-Exit
Enter Your Choice, 0 to end: 1
Input $ to break, else press enter:
Input data for node: 44
Input $ to break, else press enter:
Input data for node: 88
Input $ to break, else press enter:$
Enter Your Choice, 0 to end: 4
-->44 -->88
Enter Your Choice, 0 to end: 2
Insert the node value : 55
Enter Your Choice, 0 to end: 4
-->55 -->44 -->88
Enter Your Choice, 0 to end: 0

1-ADD
2-DELETE
3-DISPLAY
4-EXIT
Enter Your Choice: 1
Enter Item To Add: 44
Item Added is: 44
Enter Your Choice: 1
Enter Item To Add: 55
Item Added is: 55
Enter Your Choice: 3
FRONT-->44-->55<--REAR
Enter Your Choice: 2
Item Deleted Is: 44
Enter Your Choice: 3
FRONT-->55<--REAR
Enter Your Choice: 4