## Euler Problem: Solution of Problem-5 in C++

```#include

#define X 20

using namespace std;

long hcf ( const long & small_num, const long & big_num ) {
if ( small_num > big_num ) {
return hcf ( big_num, small_num );
}
else {
if ( big_num % small_num == 0 ) {
return small_num;
}
else {
return hcf ( big_num % small_num, small_num );
}
}
}

int main ( int argc, char *argv[] ) {
long result = 1;
long temp_hcf;
for ( int i = 1; i <= X; i++ ) {
temp_hcf = hcf ( result, i );
result /= temp_hcf;
result *= i;
}
cout << "Result = " << result << endl;
return 0;
}

// Result = 232792560```

## Euler Problem: Solution of Problem-4 in C++

Q) A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 * 99. Find the largest palindrome made from the product of two 3-digit numbers.

```Lets take the case: largest palindrome made from the product of two single-digit numbers.
what will be the smallest product of two single-digit numbers?
of-course the product of two smallest-single-digit numbers ie. 0 and 0 ie. 0-
and the largest product?
product of two largest-single-digit numbers ie. 9 and 9 ie. 81

Now we are interested in only palindrome numbers, that too between the above found numbers
( 0 and 81 )List of palindrome numbers between 0 and 81: 1, 11, 22, 33, 44, 55, 66, 77
Note: 88 and 99 won't get into the list.
since we are looking for largest palindrome number which will be the product of two single-digit numbers,
let's go through the list from highest to lowest and with each number in the list, we will check if it is
a product of two one digit number.
We will stop traversing through the list of numbers when we find such a number.

Now let's do a little optimization to our logic. Why do we need to generate a list of all palindrome numbers
below 81?
So first we will find the immediate palindrome number below 81, will check if it suffice our need,
if not go for next.
Better.. , isn't it?

The same logic can be scaled up to find largest palindrome made from the product of two n-digit numbers.

We make use of a function in our code to divide the task.
returns the palindrome number just below the argument.
Rather than a brute-force search I use a few heuristics here:
A palindrome number is a smaller number combined with the same number reversed.
Eg:
1221 = 12 combined with reverse of 12 ie 21
765567 = 765 combined with reverse(765)
In case of odd number of digits, it will be similar to above case but a digit in between.
Eg:
12821 = '12' + a digit(here '8') + reverse(12)
12321 = '12' + '3' + '21'
So a palindrome less than a number can be found by splitting the number into two sub-parts:
In case of even number of digits:
two numbers of half the number of digits ie 1221 into 12 and 21, 765567 into 765 and 567
In case of odd number of digits:
the small number and middle digit forms first number and reversed number forms the second
number ie 12821 into 128 and 21, 12321 into 123 and 21.
Now the first number is decremented by 1. Eg: 12 of 1221 becomes 11, 765 of 765567 becomes 764
128 of 12821 becomes 127, etc
Now we just need to take the reverse of particular digits of this new number and append to the new
number to get the required palindrome number.
ie. 	in case of even number of digits, just reverse the new number.
in case of odd number of digits, reverse all digits except the last.

Lets try our logic with a few numbers to ensure its correctness:
1) Find the palindrome number just below 139890?
This number has even number of digits ( 6 digits)
It is split into two sub-parts: 139 and 890
Now the first number is decremented by 1 and becomes 138.
Now the reverse of new number is appended with the new number  to form the final answer.
ie '138' + '831' = 138831 (Note: + doesn't mean mathematical addition)
You may have a doubt now, why the first number is reduced and not the second number.
If we reduce the second number and prepend with the reverse of new number, we wont get
the expected result. In this case (831 - 1) = 830, '038' + '830' = 038830. Is this what
we wanted?
A BIG NO... !

Now with a number of odd number of digits:
2) 1637001
yes, you are right, 7 digits
right.., split into 1637 and 001
Note: 001 wont be considered as 1
1637 is decremented by 1 and becomes 1636
Now, since odd number of digits, only 1, first 6 and 3 is reversed and appended.

Lets go for one more
3)	13300029212
11 digits.
133000 and 29212
133000 - 1 = 132999
'132999' + '99231' = 13299999231

"palindromeNumJustBelow ( long input )" implementation:
First, I insert each digit of the argument ie input into a vector. so the vector can be seen as an array
of digits. This is for my easiness to handle the digits of the number. Since I use the push_back() function
of the vector, The order of the digits will be reversed. Then I set a variable numOfDigits to number of digits
in the number. Then a new number ie 'temp_input_half' is generated with last
half of digits (Note: last half digits means first half digits of the number, don't get confused).
'temp_input_half' is decremented by 1.
The second half digits in the vector are replaced with the digits of new value of 'temp_input_half'
Now our final array of digits is ready; we just need to form the final number out of it.
I have used two lambda functions in the code. It's just a way to define a function somewhere just for one time
use. I use those with for_each loop.

At the top of the program, num_of_digits_in_num is defined as 3, ie. it should be a multiple of two 3-digit
numbers.

In the main function:
I find
smallest 3-digit number
( to avoid hard-coding, we use 'num_of_digits_in_num',
so when i say 3 assume that I used 'num_of_digits_in_num' in the code)
largest 3-digit number
Then the execution gets into a while loop which exits if the "palindrome_num" (palindrome number) is less than
the product of two smallest-3-digit-number. Product of two 3-digit-number should be between product of two
smallest 3-digit number and product of two largest 3-digit number. It's simple mathematics.

Note: first the "palindrome_num" is set to product of two largest 3-digit number before the while-loop

Inside while-loop, we find a number just below the current "palindrome_num" and we use another for-loop which
iterates from largest 3-digit number to smallest 3-digit number and in each iteration it is checked if it's
possible to divide the palindrome number with the iteration variable (in our code, "temp_div1") and still get
a 3-digit number

Largest palindrome made from the product of two 3-digit numbers is 906609 = 993 * 913

Code for reference:

#include <iostream>
#include <cmath>
#include <algorithm>
#include <iterator>
#include <vector>
#include <boost/lambda/lambda.hpp>
#include <boost/functional.hpp>
#include <boost/function.hpp>

#define num_of_digits_in_num 3

using namespace std;
using namespace boost::lambda;

long palindromeNumJustBelow ( long input ) {
vector inputAsArray; //array of single-digit-numbers

long temp = input;
while ( temp ) {
inputAsArray.push_back ( temp%static_cast(10) );
temp = floor( temp/static_cast(10) );
}

int numOfDigits = inputAsArray.size();

temp = floor ( ( inputAsArray.size()/2 ) ) ;
long temp_input_half = 0;
boost::function f2 = ( var(temp_input_half)*=10, var(temp_input_half)+=boost::lambda::_1);
for_each ( inputAsArray.rbegin(), inputAsArray.rend()-temp, f2 );

temp_input_half--;

long temp1 = temp;
while ( temp_input_half ) {
inputAsArray [ temp1++ ] = ( temp_input_half%static_cast(10) );
temp_input_half = floor( temp_input_half/static_cast(10) );
}

// mirror all the digits from index temp till end
while ( temp < numOfDigits ) {
inputAsArray [ numOfDigits - temp -1 ] = inputAsArray [ temp ] ;
temp ++ ;
}

temp = 0;
boost::function f1 = ( var(temp)*=10, var(temp)+=boost::lambda::_1);

for_each ( inputAsArray.rbegin(), inputAsArray.rend(), f1 );

return temp;
}
int main ( int argc, char * argv[] ) {

long num_min, num_max;
if ( num_of_digits_in_num==1 ) {
num_min = 0;
}
else {
num_min = pow( static_cast(10), static_cast(num_of_digits_in_num-1) );
}

num_max = (pow( static_cast(10), static_cast(num_of_digits_in_num) ) - 1 );

long prod_min, prod_max;
prod_min = num_min*num_min;
prod_max = num_max*num_max;

long palindrome_num = prod_max;
long temp_div2;

bool ans_found = false;
while ( palindrome_num > prod_min && !ans_found) {

for (int temp_div1 = num_max; temp_div1 >= num_min && !ans_found; temp_div1-- ) {
temp_div2 = palindrome_num/temp_div1;
if ( palindrome_num % temp_div1 == 0 && temp_div2 <= num_max && temp_div2 >= num_min ) {
cout << "Largest palindrome made from the product of two "
<< num_of_digits_in_num << "-digit numbers is " << palindrome_num
<< " = " << temp_div1 << " * " << temp_div2 << endl;
ans_found = true;
}

}
}
}```

## Euler Problem: Solution of Problem-3 in C++

```#include <algorithm>
#include <iostream>
#include <vector>
#include <math.h>

using namespace std;

unsigned long largest_prime_factor_of ( unsigned long num ) {
clog << "largest_prime_factor_of(" << num << ")" << endl;
unsigned long largest_prime_divisor = -1;
if ( num < 2 ) {
return -1;
}
else if ( num >=2 && num<=3 ) {
return num;
}
while ( num%static_cast<unsigned long>(2) == 0 ) {
largest_prime_divisor = 2;
cout << "/2" << endl;
//cout << "num%static_cast(2) = " << num%static_cast(2) << " num/static_cast(2) = " << num/static_cast(2) << endl;
num /= 2;
}
while ( num%3 == 0 ) {
largest_prime_divisor = 3;
cout << "/3" << endl;
//cout << "num%static_cast(3) = " << num%static_cast(3) << " num/static_cast(3) = " << num/static_cast(3) << endl;
num /= 3;
}

vector<unsigned long> prime_num_vector;
prime_num_vector.push_back(static_cast<unsigned long>(3));

clog << "largest_prime_divisor(after checking till 3) = " << largest_prime_divisor << "and num = " << num << endl;

unsigned long divisor = 3;
bool divisor_is_prime;
while ( (divisor+2) <= num ) {
//cout << ".";
divisor += 2;
clog << "divisor = " << divisor << endl;
divisor_is_prime=true;
double sqrt_of_divisor = sqrt(divisor);
clog << "sqrt_of_divisor = " << sqrt_of_divisor << endl;
clog << "going inside for-loop" << endl;
for( vector::iterator iter = prime_num_vector.begin();
iter != prime_num_vector.end();
iter++)
{
clog << "\t*iter = " << *iter << endl;
if ( *iter > sqrt_of_divisor ) {
clog << "\t*iter > sqrt_of_divisor" << endl;
break;
}
if ( divisor % *iter == 0 ) {
clog << "\tdivisor_is_prime = false" << endl;
divisor_is_prime = false;
break;
}
}
if ( divisor_is_prime ) {
clog << "divisor_is_prime (" << divisor << ")" << endl;
prime_num_vector.push_back ( divisor );

while ( num % divisor == 0 ) {
largest_prime_divisor = divisor;
num /= divisor;
cout << "largest_prime_divisor changed to " << largest_prime_divisor << " and num = " << num << endl;
}
}
}
clog << "returning " << largest_prime_divisor << endl;
return largest_prime_divisor;
}

int main ( int argc, char *argv[] ) {
unsigned long l = 600851475143 ;
//unsigned long l = 12 ;
cout << endl << "largest_prime_factor_of(l) = " << largest_prime_factor_of(l) << endl;
}```

## Euler Problem: Solution of Problem-2 in C++

Q) Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

# Reference:

```#include <iostream>

#define MAX 4000000

using namespace std;
/*
Series: 1 1 2 3 5 8 13 21 34 55 89 144 ...
Series we are interested in: 2 8 34 144 ...
we can see a relation for the above series:
f(0) = 2
f(1) = 8
f(n) = 4*f(n-1) + f(n-2)
Now we dont need any check for even/odd
*/
int main ( int argc, char *argv[] ) {
unsigned long var_a = 2 ;
unsigned long var_b = 8 ;
unsigned long sum = 0 ;
unsigned long temp = 0 ;
while ( var_b < MAX ) {
sum += var_b ;
//cout << endl;
temp = var_b ;
var_b = ( 4 * var_b + var_a ) ;
var_a = temp ;
}
cout << sum << endl;
}```

## Ball in the Box (A C++&OpenGL game)

Ball in the Box is a simple game developed in C++ using OpenGL library during my free time, out of my interest to learn OpenGL. The latest version of it is 4.0. Since it is not in my present system, I will give you the 3.0 version’s source code. You are free to modify or copy or what-ever you like to do with this… 🙂

Before going into the code, let me give you a feel of my game:

/*

File name : DisplayText.h

Author : Sreejith Sreekantan

Version : 3.0

Date : June 2012

*/

#ifndef __DisplayText_h__

#define __DisplayText_h__

#ifdef __APPLE__

#include <GLUT/glut.h>

#include <OpenGL/OpenGL.h>

#include <OpenGL/gl.h>

#else

#include <GL/glut.h>

#endif

#include

using namespace std;

namespace DisplayText {

void DisplayText(float posx, float posy, void* font, char string[]) {

glRasterPos2f(posx, posy);

while(*string) {

glutBitmapCharacter(font, *string);

string++;

}

}

void DisplayText(float posx, float posy, void* font, int value) {

int iTemp1=0, iTemp2 = value;

while(iTemp2>0) {

iTemp1++;

iTemp2 /= 10;

}

char *string = (char*)malloc(sizeof(char)*iTemp1+1);

sprintf(string, “%d”, value);

//clog << “string=” << *string << endl;

DisplayText(posx, posy, font, string);

}

}

#endif // __DisplayText_h__

/*

File name : BitB (Ball in the Box)

Version : 3.0

Date : June 2012

*/

#ifdef __APPLE__

#include <GLUT/glut.h>

#include <OpenGL/OpenGL.h>

#include <OpenGL/gl.h>

#else

#include <GL/glut.h>

#endif

#include “DisplayText.h”

#include

#include

#include

#include

using namespace std;

#define NAME_OF_GAME “Ball in the Box!”

// Game states

#define GAME_INIT 0

#define GAME_START 1

#define GAME_END 2

#define GAME_CLOSE 3

#define SCORE_OFFSET 5

#define SCREEN_WIDTH_PIX 120

#define SCREEN_HEIGHT_PIX 80

#define DASHBOARD_WIDTH_PIX SCREEN_WIDTH_PIX

#define DASHBOARD_HEIGHT_PIX 10

#define BALLSPEED_OFFSET 0.05

#define BOARD_WIDTH 5.0

#define BOARD_LENGTH 20.0

#define BOARD_HEIGHT 3

#define PLAYER_NAME_LEN 11

int iGameStatus;

int iScore;

int iScoreTemp=0;

GLfloat fBallSpeed = 0.01;

GLfloat fBoardSpeed = 4.0;

GLfloat afBallSpecular[] = {0.5, 0.0, 0.0, 1.0};

GLfloat afBallAmbient[] = {0.5, 0.0, 0.0, 1.0};

GLfloat afBallShininess[] = {50.0};

GLfloat fBallTransX = SCREEN_WIDTH_PIX/2;

GLfloat fBallTransY = SCREEN_HEIGHT_PIX/2;

GLfloat fBallTransXOffset = fBallSpeed;

GLfloat fBallTransYOffset = fBallSpeed;

GLfloat fBoardTransX = 0.0;

GLfloat fBoardTransY = 0.0;

GLfloat fBoardTransXOffset = fBoardSpeed;

char acPlayerName[PLAYER_NAME_LEN];

GLuint iSphereGenList;

GLuint iBoardGenList;

GLuint iDashboardGenList;

GLuint iGetNameGenList;

GLuint iShowScore;

time_t startTime;

void fnDisplay() {

gluLookAt(0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0);

glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);

if (iGameStatus == GAME_INIT){

glPushMatrix();

glCallList(iGetNameGenList);

glColor3f(0.75, 0.0, 0.0);

DisplayText::DisplayText(SCREEN_WIDTH_PIX/2-8, SCREEN_HEIGHT_PIX/2, GLUT_BITMAP_HELVETICA_18, acPlayerName);

glPopMatrix();

}

else if (iGameStatus == GAME_START) {

iScore = ((time(NULL) – startTime)/SCORE_OFFSET);

if(iScore > 0 && iScore % (SCORE_OFFSET*2) == 0 && iScore != iScoreTemp) {

fBallTransX *= 1.025;

fBallTransY *= 1.025;

fBoardTransX *= 1.025;

fBoardTransY *= 1.025;

iScoreTemp = iScore;

clog << “Speed incremented:” << fBallTransX << “,” << fBallTransY << endl;

}

//clog << “iScore=” << iScore << endl;

glPushMatrix();

glTranslatef(fBallTransX, fBallTransY, 0.0);

glCallList(iSphereGenList);

glPopMatrix();

glPushMatrix();

glTranslatef(fBoardTransX, fBoardTransY, 0.0);

glCallList(iBoardGenList);

glPopMatrix();

glPushMatrix();

glCallList(iDashboardGenList);

glColor3f(1.0, 1.0, 1.0);

DisplayText::DisplayText(10.0, SCREEN_HEIGHT_PIX+5, GLUT_BITMAP_HELVETICA_18, acPlayerName);

DisplayText::DisplayText(SCREEN_WIDTH_PIX-10, SCREEN_HEIGHT_PIX+5, GLUT_BITMAP_HELVETICA_12, iScore);

glPopMatrix();

}

else if (iGameStatus == GAME_END){

glPushMatrix();

glCallList(iShowScore);

glColor3f(0.75, 0.0, 0.0);

DisplayText::DisplayText(SCREEN_WIDTH_PIX/2-8, SCREEN_HEIGHT_PIX/2, GLUT_BITMAP_HELVETICA_18, iScore);

glPopMatrix();

}

else if (iGameStatus == GAME_CLOSE){

exit(0);

}

glFlush();

}

void fnReshape(int w, int h) {

glViewport(0, 0, w, h);

glMatrixMode(GL_PROJECTION);

glOrtho(0, SCREEN_WIDTH_PIX, 0, SCREEN_HEIGHT_PIX+DASHBOARD_HEIGHT_PIX, 0.5, 2);

glMatrixMode(GL_MODELVIEW);

}

void fnIdle() {

//clog << “fnIdle():” << endl;

bool flag_x_offset_changed = false;

bool flag_y_offset_changed = false;

if (fBallTransX == SCREEN_WIDTH_PIX || fBallTransX == 0) {

fBallTransXOffset *= -1; // inverse x direction

flag_x_offset_changed = true;

}

if (fBallTransY == SCREEN_HEIGHT_PIX) {

fBallTransYOffset *= -1; // inverse y direction

flag_y_offset_changed = true;

}

(fBallTransX-BALL_RADIUS) <= (fBoardTransX + BOARD_LENGTH) &&

(fBallTransY) >= BOARD_HEIGHT) // if ball is hit by the board

{

fBallTransYOffset *= -1; // inverse y direction

iScore += SCORE_OFFSET;

flag_y_offset_changed = true;

}

if (fBallTransY == 0) {

glutIdleFunc(NULL);

iGameStatus = GAME_END;

clog << “Game status changed to GAME_END” << endl;

}

fBallTransX += fBallTransXOffset;

fBallTransY += fBallTransYOffset;

if (!flag_x_offset_changed) {

if (fBallTransX > SCREEN_WIDTH_PIX) {

fBallTransX = SCREEN_WIDTH_PIX;

}

else if (fBallTransX < 0) {

fBallTransX = 0.0;

}

}

if (!flag_y_offset_changed) {

if (fBallTransY > SCREEN_HEIGHT_PIX) {

fBallTransY = SCREEN_HEIGHT_PIX;

}

else if (fBallTransY < 0) {

fBallTransY = 0.0;

}

}

glutPostRedisplay();

}

void fnSpecial(int key, int x, int y) {

//clog << “fnSpecial():” << “key=” << key << ” x=” << x << ” y=” << y << endl;

switch (key) {

case 100: // left arrow key

if (fBoardTransX > 0.0 && iGameStatus == GAME_START) {

fBoardTransX -= fBoardTransXOffset;

}

break;

case 102: // right arrow key

if ( (fBoardTransX + BOARD_LENGTH) < SCREEN_WIDTH_PIX && iGameStatus == GAME_START) {

fBoardTransX += fBoardTransXOffset;

}

break;

//case 101: // up arrow key

//case 103: // down arrow key

default:

break;

}

glutPostRedisplay();

}

int iTemp =0;

void fnKeyboard(unsigned char key, int x, int y) {

clog << “fnKeyboard():” << “key=” << key << “(int)key=” << (int)key << ” x=” << x << ” y=” << y << endl;

switch (key) {

case 13: // if enter key pressed

if ( acPlayerName[0] != ” && iGameStatus == GAME_INIT) {

fBallTransX = SCREEN_WIDTH_PIX/2;

fBallTransY = SCREEN_HEIGHT_PIX/2;

fBallTransXOffset = fBallSpeed;

fBallTransYOffset = fBallSpeed;

glutIdleFunc(fnIdle);

iScore = 0;

startTime = time(NULL);

clog << “startTime=” << startTime << endl;

iGameStatus = GAME_START;

clog << “Game status changed to GAME_START” << endl;

}

else if (iGameStatus == GAME_END) {

iGameStatus = GAME_CLOSE;

clog << “Game status changed to GAME_CLOSE” << endl;

}

break;

case 8: // if backspace key pressed

if (iTemp >= 0 && iGameStatus == GAME_INIT) {

acPlayerName[–iTemp] = ”;

}

break;

case 27: // if escape key is pressed

if (iGameStatus == GAME_END) {

iGameStatus = GAME_INIT;

fnKeyboard(13, x, y);

return;

}

default:

if ( (key >= ‘a’ || key >= ‘A’) &&

(key <= ‘z’ || key <= ‘Z’) &&

iTemp < PLAYER_NAME_LEN-1 &&

iGameStatus == GAME_INIT

)

{

acPlayerName[iTemp++] = key;

acPlayerName[iTemp] = ”;

clog << “iTemp=” << iTemp << endl;

}

break;

}

glutPostRedisplay();

}

void init() {

//clog << “init():” << endl;

iGameStatus = GAME_INIT;

GLfloat light0_position[] = {-1.0, -1.0, -1.0, 0.0};

glLightfv(GL_LIGHT0, GL_POSITION, light0_position);

iBoardGenList = glGenLists(1);

glNewList(iBoardGenList, GL_COMPILE);

glEnable (GL_LINE_STIPPLE);

glLineStipple (1, 0x0F0F);

glColor3f(0.5, 0.75, 0.75);

glBegin(GL_LINES);

glVertex2d(0.0, BOARD_HEIGHT);

glVertex2d(BOARD_LENGTH, BOARD_HEIGHT);

glEnd();

glDisable(GL_LINE_STIPPLE);

glEndList();

iSphereGenList = glGenLists(1);

glNewList(iSphereGenList, GL_COMPILE);

glEnable(GL_LIGHTING);

glEnable(GL_LIGHT0);

glMaterialfv(GL_FRONT, GL_SPECULAR, afBallSpecular);

glMaterialfv(GL_FRONT, GL_AMBIENT, afBallAmbient);

glMaterialfv(GL_FRONT, GL_SHININESS, afBallShininess);

glDisable(GL_LIGHT0);

glDisable(GL_LIGHTING);

glEndList();

iDashboardGenList = glGenLists(1);

glNewList(iDashboardGenList, GL_COMPILE);

glColor3f(0.0, 0.5, 0.5);

glBegin(GL_POLYGON);

glVertex2d(0, SCREEN_HEIGHT_PIX);

glVertex2d(0, SCREEN_HEIGHT_PIX+DASHBOARD_HEIGHT_PIX);

glVertex2d(SCREEN_WIDTH_PIX, SCREEN_HEIGHT_PIX+DASHBOARD_HEIGHT_PIX);

glVertex2d(SCREEN_WIDTH_PIX, SCREEN_HEIGHT_PIX);

glEnd();

glColor3f(1.0, 1.0, 1.0);

DisplayText::DisplayText(1.0, SCREEN_HEIGHT_PIX+5, GLUT_BITMAP_HELVETICA_12, “Player:”);

DisplayText::DisplayText(SCREEN_WIDTH_PIX-20, SCREEN_HEIGHT_PIX+5, GLUT_BITMAP_HELVETICA_12, “Score:”);

glEndList();

iGetNameGenList = glGenLists(1);

glNewList(iGetNameGenList, GL_COMPILE);

glColor3f(0.0, 0.5, 0.5);

glBegin(GL_POLYGON);

glVertex2d(20, SCREEN_HEIGHT_PIX/2-20);

glVertex2d(20, SCREEN_HEIGHT_PIX/2+20);

glVertex2d(SCREEN_WIDTH_PIX-20, SCREEN_HEIGHT_PIX/2+20);

glVertex2d(SCREEN_WIDTH_PIX-20, SCREEN_HEIGHT_PIX/2-20);

glEnd();

glColor3f(0.8, 0.8, 0.8);

DisplayText::DisplayText(23.0, SCREEN_HEIGHT_PIX/2+10, GLUT_BITMAP_HELVETICA_18,

“Enter your name and press enter to continue.”);

glBegin(GL_POLYGON);

glVertex2d(SCREEN_WIDTH_PIX/2-20, SCREEN_HEIGHT_PIX/2-7);

glVertex2d(SCREEN_WIDTH_PIX/2-20, SCREEN_HEIGHT_PIX/2+7);

glVertex2d(SCREEN_WIDTH_PIX/2+20, SCREEN_HEIGHT_PIX/2+7);

glVertex2d(SCREEN_WIDTH_PIX/2+20, SCREEN_HEIGHT_PIX/2-7);

glEnd();

glEndList();

iShowScore= glGenLists(1);

glNewList(iShowScore, GL_COMPILE);

glColor3f(0.0, 0.5, 0.5);

glBegin(GL_POLYGON);

glVertex2d(20, SCREEN_HEIGHT_PIX/2-20);

glVertex2d(20, SCREEN_HEIGHT_PIX/2+20);

glVertex2d(SCREEN_WIDTH_PIX-20, SCREEN_HEIGHT_PIX/2+20);

glVertex2d(SCREEN_WIDTH_PIX-20, SCREEN_HEIGHT_PIX/2-20);

glEnd();

glColor3f(0.8, 0.8, 0.8);

DisplayText::DisplayText(30.0, SCREEN_HEIGHT_PIX/2+10, GLUT_BITMAP_HELVETICA_18,

DisplayText::DisplayText(23.0, SCREEN_HEIGHT_PIX/2-17, GLUT_BITMAP_HELVETICA_18,

“Press enter to exit or escape to continue..”);

glBegin(GL_POLYGON);

glVertex2d(SCREEN_WIDTH_PIX/2-20, SCREEN_HEIGHT_PIX/2-7);

glVertex2d(SCREEN_WIDTH_PIX/2-20, SCREEN_HEIGHT_PIX/2+7);

glVertex2d(SCREEN_WIDTH_PIX/2+20, SCREEN_HEIGHT_PIX/2+7);

glVertex2d(SCREEN_WIDTH_PIX/2+20, SCREEN_HEIGHT_PIX/2-7);

glEnd();

glEndList();

glLineWidth(BOARD_WIDTH);

glClearColor(0.5, 0.5, 0.5, 1.0);

glEnable(GL_BLEND);

glBlendFunc (GL_SRC_ALPHA, GL_ONE_MINUS_SRC_ALPHA);

}

int main(int argc, char **argv) {

glutInit(&argc, argv);

glutInitDisplayMode(GLUT_SINGLE);

glutInitWindowSize(600, 400);

glutInitWindowPosition(100, 100);

glutCreateWindow(NAME_OF_GAME);

init();

glutDisplayFunc(fnDisplay);

glutReshapeFunc(fnReshape);

//glutIdleFunc(fnIdle);

glutSpecialFunc(fnSpecial);

glutKeyboardFunc(fnKeyboard);

glutMainLoop();

}

Not a big deal.. I know.. but a beginner in OpenGL will find it useful.. 🙂

## ‘const’ keyword and pointers in C++ (CPP) language

Many a times one of the challenges a beginner faces is with the use of const keyword. It comes handy when one understands the exact use of it. Const keyword changes its meaning depending on where it is being used. Below I give all possible use of the keyword.

Normal integer variable:
int intValue                                                                            // integer variable
As function parameter:
Void testFunc(int intValue) {
// modification of intValue is possible
intValue = 10;
}
Here the value passed to testFunc() is copied to a separate memory area, which can be accessed through intValue

Pointer to integer variable:
int *ptrToInt                                                              // pointer to integer variable
As function parameter:
void testFunc(int *ptrToInt) {
int a = 10;
cout << ptrToInt << endl;               //prints the memory address of passed variable
cout << *ptrToInt << endl;

ptrToInt = &a;                                               //assigning address of variable ‘a’ to ptrToInt ie. Modification of ptrToInt is possible

cout << ptrToInt << endl;               //prints the memory address of variable a
cout << *ptrToInt << endl;  //prints the value of a ie. 10
}

Constant integer value:
int const constInt                                                      // constant variable, cant be modified
const int constInt                                                      // also
As function parameter:
void testFunc(int const constInt) {
//constInt can’t be changed here
}

Pointer to constant integer:
int const *ptrToConstInt                                          // pointer to a constant integer variable
const int *ptrToConstInt                                          // also
As function parameter:
void testFunc(const int *ptrToConstInt) {
const int a = 10;
ptrToConstInt = &a;
cout << ptrToConstInt << endl;               // prints the memory address of variable a
cout << *ptrToConstInt << endl;            // prints the value of a ie. 10
}

Constant pointer to an integer:
int *const constPtrToInt                                           // constant pointer to integer variable
As function parameter:
void testFunc(int *const constPtrToInt) {
//constPtrToInt cant be changed but the value it points to, can be altered
*constPtrToInt += 10;                 //since it points to, is a normal integer value
}

Constant pointer to constant integer:
int const * const constPtrToConstInt          // constant pointer to constant integer value
As function parameter:
void testFunc(int const *const constPtrToConstInt) {
//here the value of constPtrToConstInt as well as the value it points to, both is immutable
}

MORE USE OF ‘const’ KEYWORD:

Constant pointer to a constant pointer to an integer:
int * const * const const_ptr_to_const_ptr_to_int

Constant pointer to a pointer to a pointer:
int ** const const_ptr_to_ptr_to_int

Pointer to a constant pointer to an integer:
int * const * ptr_to_const_ptr_to_int

Pointer to a pointer to an integer:
int ** ptr_to_ptr_to_int

Tip: “int const” and “const int” are same. So if the line contains any of these two, understand that the final part is “integer constant” take the rest of the line and read backwards,
*                     ==means==>         ‘pointer to’
* const            ==means==>         ‘constant pointer to’
Multiple ‘*’        ==means==>         that number of ‘pointer to’

Hope this post was helpful to you… 🙂