Create a Tic tac toe game using c++

By -
The following program is the source code of a game called tic tac toe. The program is implemented using c++.

Code : 


#include<iostream>
#include<windows.h>
using namespace std;
char a[9]={'1','2','3','4','5','6','7','8','9'};
char c1,c2;
void draw_board()
{
    cout<<"\n\n";
    cout<<"   "<<a[0]<<"   |   "<<a[1]<<"   |    "<<a[2]<<"   ";
    cout<<endl<<"       |       |       ";
    cout<<"\n-------|-------|-------";
    cout<<endl<<"   "<<a[3]<<"   |   "<<a[4]<<"   |    "<<a[5]<<"   ";
    cout<<endl<<"       |       |       ";
    cout<<"\n-------|-------|-------";
    cout<<endl<<"   "<<a[6]<<"   |   "<<a[7]<<"   |    "<<a[8]<<"   ";
    cout<<endl<<"       |       |       ";

}
int check_res()
{
    if(a[0]==a[1] && a[1]==a[2])
    {
        if(a[0]==c1)
            return 1;
        else
            return 2;
    }

    else if(a[0]==a[3] && a[3]==a[6])
    {
        if(a[0]==c1)
            return 1;
        else
            return 2;
    }
    else if(a[0]==a[4] && a[4]==a[8])
    {
        if(a[0]==c1)
            return 1;
        else
            return 2;
    }
    else if(a[1]==a[4] && a[4]==a[7])
    {
        if(a[1]==c1)
            return 1;
        else
            return 2;
    }
    else if(a[2]==a[4] && a[4]==a[6])
    {
        if(a[2]==c1)
            return 1;
        else
            return 2;
    }
    else if(a[2]==a[5] && a[5]==a[8])
   {
        if(a[2]==c1)
            return 1;
        else
            return 2;
    }
    else if(a[3]==a[4] && a[4]==a[5])
    {
        if(a[3]==c1)
            return 1;
        else
            return 2;
    }
    else if(a[6]==a[7] && a[7]==a[8])
    {
        if(a[6]==c1)
            return 1;
        else
            return 2;
    }
    else
        return 0;

}
void play_game()
{

    int c,r;
    int p;
    cout<<"\nEnter your choice : \n1.X\t 2.O\n";
    cin>>c;
    if(c==1)
    {
        c1='X';
        c2='O';
    }
    else if(c==2)
    {
        c1='O';
        c2='X';
    }
    else
    {
        cout<<"\nInvalid choice...";

    }
    c=0;
    while(c<10)
    {
        A:
        cout<<"\n==>Player 1 turn :";
        cout<<"\nEnter your choice : ";
        cin>>p;
        if(isdigit(a[p-1]))
        {
            c++;
            a[p-1]=c1;
            system("cls");
            draw_board();
            r=check_res();
            if(r!=0)
                break;

        }
        else
        {
            cout<<"\nAlready filled up";
            goto A;
        }
        B:
        cout<<"\n==>Player 2 turn :";
        cout<<"\nEnter your choice : ";
        cin>>p;
        if(isdigit(a[p-1]))
        {
            a[p-1]=c2;
            system("cls");
            draw_board();
            r=check_res();
            c++;
            if(r!=0)
                break;

        }
        else
        {
            cout<<"\nAlready filled up";
            goto B;
        }
    }
    if(r!=0)
            cout<<"\nPlayer "<<r<<" wins the match";
    else
        cout<<"\nDraw";

}

int main()
{
    draw_board();
    play_game();
    return 0;
}

Output :




Comments

Post a Comment

Popular posts from this blog

Non Restoring Division Algorithm Implementation in C

By Sidharth Patnaik -
Image
This  is a Dynamic program for the implementation of the NON RESTORING Division Algorithm in C Language. Non Restoring Division uses Left Shift Operations, Twos' Compliment and Binary Addition. Screenshots[Demo] The Code #include<stdio.h> #include<malloc.h> int *a,*q,*m,*mc,*c,n,d; int powr(int x,int y) { int s=1,i; for(i=0;i<y;i++) s=s*x; return s; } void print(int arr[],int n) { int i; for(i=0;i<n;i++) printf("%d ",arr[i]); } void bin(int n, int arr[]){ int r, i = 0; do{ r = n % 2; n /= 2; arr[i] = r; i++; }while(n > 0); } void set(int array[], int x){ int i,tmp[20]={0}; for(i = x -1; i >=0; i--) tmp[x-1-i]=array[i]; for(i=0;i<x;i++) array[i]=tmp[i]; } int len(int x) { int i=0; while(powr(2,i)<=x) i++; return ++i; } void addBinary(int a1[], int a2[]) { int bi[2]={0},ca[20]={0}; int t=len(n),tmp=0; int *su=(int*)malloc(sizeof(int)*len(n)); while(t...
Read more

Hackerrank Modified Kaprekar Numbers Solution

By Sidharth Patnaik -
Image
This is the solution to the Modified Kaprekar Numbers found in the implementation section of the algorithm domain in Hackerrank. A modified Kaprekar number is a positive whole number n with d digits, such that when we split its square into two pieces - a right hand piece r with d digits and a left hand piece l that contains the remaining d or d−1 digits, the sum of the pieces is equal to the original number (i.e. l + r = n) . Screenshot The Code def sendKap(x,s): y=x*x y=str(y) s1=y[:len(y)/2:] s2=y[len(y)/2::] s3=int(s1)+int(s2) if(s3==x): s.append(str(x)) st=input() en=input() s=[] for i in range(st,en+1): if(i>9): sendKap(i,s) elif(i==1): s.append(str(i)) elif(i==9): s.append(str(i)) if(len(s)==0): print 'INVALID RANGE' else: print ' '.join(s) Got any problems feel free to comment them here !
Read more

Bit Stuffing Code Implementation in Java

By -
Image
This is an implementation of the Bit Stuffing popularly used in data communication, also known as one of the concepts of Framing data bits. To know more please  click here. Sample Output: This is a sample output with data binary data entered as 11001111110. Program Code: import java.util.*; public class BitStuffing { public static void main(String[] args) { System.out.print("Enter the Binary message: "); Scanner sn=new Scanner(System.in); String data = sn.nextLine(); String res = new String(); String out=new String(); int counter = 0; for(int i=0;i<data.length();i++) { if (data.charAt(i)!='1' && data.charAt(i)!='0') { System.out.println("Enter only Binary values!!!"); return; } ...
Read more