Thread: Using C programming language, design and develop a two-player number game which can b

Sounds a lot like the Game of Nim....

(Sorry for not being of any help here)

Originally Posted by grumpy

My take is that you have a bit of code to start meeting a couple of the dozen requirements, but you haven't really met any of the requirements as yet. You need to work systematically through the requirements, rather than simply posting here in the hope that someone else will complete your work for you.

I'm ready to work on it, but believe me I'm short of some right resources, if you can please help me out with appropriate/similar/helping example links, I'll get to work because Google did not help me much even after long hours of searching

I'll get to work because Google did not help me much even after long hours of searching

O_o

Your internet must be broken.

grumpy is saying, in his way, that we aren't going to do this for you. We will help, but we aren't going to do it for you.

So, taking the advice: what does "The player who pops the last balloon is the winner of the game." mean in code?

Soma

Code:

#include <stdio.h>
#include <stdlib.h>


int main()
{
    char player1[20],player2[20];
    int turn,x=10,num,j;
    printf("\t\nPlease enter name of player who will take the first turn:");
    scanf("%s",&player1);
    printf("\n %s will take the first turn",player1);
    printf("\n\nPlease enter name of player who will take the second turn:");
    scanf("%s",&player2);
     printf("\n %s will take the second turn",player2);
      printf("\n\t\nGame started");
     printf("\t\n\t\nOOOOOOOOOO");
     while(x<=10)
{






     printf("\n\n\t%s, your turn",player1);
     printf("\n\n\tEnter 1 or 2:");
     scanf("%d",&num);
     if(num>2)
     {
         printf(" Invalid choice, please try again !!");
     }




     else{
           x=x-num;




     for ( j = 1; j <= x; ++j )
     printf( "%s", "O" );


     if(x==0)
     {




         printf("The last player won!!");
         exit(0);
     }




     }


}
}

This is some revised code I could try...

I just need some help/advice so as how do I print the number of balloons popped by each player for which this code would not be much of use

And then how do I print the player's name who won!!

"the player who pops the last balloon is the winner of the game." in means x comes to 0 finally and there are no more balloons left out of 10

Originally Posted by christop

Sounds a lot like the Game of Nim....

(Sorry for not being of any help here)

Yeah, both look like cousins

Welcome to the forum, Ian!

Before you start programming, it's very helpful to sit down and do the puzzle/game/assignment by hand, without a computer.

Note the patterns you use - those are the basics of the logic for the program, since the computer will need to do the same things you did.

In a two player game, you've correctly seen that the entire game is set inside a while() loop (but a for loop or do while loop can be used).

And straight up, STOP USING those single meaningless variable names! For a simple counter, i is fine and n for a number, but once you start running amok with a, b, c, x, etc., you'll be obscuring the logic you need.

So:

Code:

int balloons = 10
int player = 1  //player 0 will play first - this is a subtlety
char players[2][30]; //holds both players names

will be some of the variables you'll want to use.
you'll want to get the players names, before the game loop starts
and store their names in players[0] and players[1].

Note that there is NO players[2], since valid index are 0 to size -1.

Code:

//start of the game loop
while(balloons>0)  {
    //if player is 0, 
        //make it 1 
    //else 
       //make it 0

    //print out how many balloons are in play
    //print out the players name on turn
    //get their number of balloons to be popped
    //and subtract them

}

announce the winner

That's a start. You WILL find subtleties as you code and test the program.

Try to make your code as CLEAR and SIMPLE as you can. If you use variable names that obscure their meaning, or write complex code, you're just making it more difficult.

Originally Posted by Adak

Welcome to the forum, Ian!

Before you start programming, it's very helpful to sit down and do the puzzle/game/assignment by hand, without a computer.

Note the patterns you use - those are the basics of the logic for the program, since the computer will need to do the same things you did.

In a two player game, you've correctly seen that the entire game is set inside a while() loop (but a for loop or do while loop can be used).

And straight up, STOP USING those single meaningless variable names! For a simple counter, i is fine and n for a number, but once you start running amok with a, b, c, x, etc., you'll be obscuring the logic you need.

So:

Code:

int balloons = 10
int player = 1  //player 0 will play first - this is a subtlety
char players[2][30]; //holds both players names

will be some of the variables you'll want to use.
you'll want to get the players names, before the game loop starts
and store their names in players[0] and players[1].

Note that there is NO players[2], since valid index are 0 to size -1.

Code:

//start of the game loop
while(balloons>0)  {
    //if player is 0, 
        //make it 1 
    //else 
       //make it 0

    //print out how many balloons are in play
    //print out the players name on turn
    //get their number of balloons to be popped
    //and subtract them

}

announce the winner

That's a start. You WILL find subtleties as you code and test the program.

Try to make your code as CLEAR and SIMPLE as you can. If you use variable names that obscure their meaning, or write complex code, you're just making it more difficult.

Thanks a ton for the idea and advice, that was really helpful,I'll keep the points in mind and try to be more organized

Will post revised/complete game once I crack it!!

Sorry for posting the solution after so many days..I had cracked it the same day, but some other assignments kept me busy

So here goes the code

Code:

#include <stdio.h>
#include <stdlib.h>


int main()
{
    char player1[20],player2[20];
    int current_player=player1;
    int balloons_play1=0;
    int balloons_play2=0;
	char dummy[5];
    int x=10,num,j;
    printf("\n\n\t\t\t *Welcome to the Balloon Pop Game*");
    printf("\n\n\ About: This is a fun game between two players which will challenge\n\tyour logic and strategy");
    printf("\n\n\ Instructions:\n\n\t *Each player can pop 1 or 2 balloons in his turn");
    printf("\n\n\t *The Player who pops the last balloon wins");


    printf("\n\n *Please enter name of player who will take the first turn:");
    scanf("%s",&player1);
    printf("\n *%s will take the first turn",player1);
    printf("\n\n *Please enter name of player who will take the second turn:");
    scanf("%s",&player2);
     printf("\n *%s will take the second turn",player2);
      printf("\n\n\n\t\t\t\t***Game started***");
     printf("\n\n\t\t\tOOOOOOOOOO balloons remaining");
     do{
{




     printf("\n\n\t*%s, your turn",current_player);
     printf("\n\n\t*Enter '1' to pop one balloon or '2' to pop 2 balloons:");
     scanf("%d",&num);


     if(num>2)
     {
         printf(" \n\n\t*Invalid choice, please try again !!");
     }




     else{
           x=x-num;




     for ( j = 1; j <= x; ++j )
     printf( "\n\t\t\t%s", "O" );
     printf("\n\n\t\t\t* %d balloons remaining *",x);


     if(current_player==player1)
     {
         balloons_play1=balloons_play1+num;
         current_player=player2;


     }
     else
     {
         balloons_play2=balloons_play2+num;
         current_player=player1;
		 getchar();
     }
     }
}
     }
	 
while(x>0);


        if(current_player==player1)
        {
            printf("\n\n Congratualations %s you won by popping the last balloon !!!",player2);


        }
        else
        {
            printf("\n\n\tCongratualations %s you won by popping the last balloon !!!",player1);
        }
         printf("\n\n\tBalloons popped by %s:%d",player1,balloons_play1);
         printf("\n\n\tBalloons popped by %s:%d",player2,balloons_play2);
		 getchar();
		 fflush(stdin);
		 printf("");
		 scanf("%s",dummy);
		
		




          
     }

Well done, Ian. Looks like you got it. You will see a benefit as your indentation style improves.

Originally Posted by Adak

Well done, Ian. Looks like you got it. You will see a benefit as your indentation style improves.

Yes, I hope to improve in time..Thank you!!

Originally Posted by ian212

Sorry for posting the solution after so many days..I had cracked it the same day, but some other assignments kept me busy

Have you already tried to compile your code with all warnings on?
You'll get a bunch of warnings mostly because of this:

Code:

char player1[20],player2[20];
int current_player=player1;

"player1" (and "player2") is an array of char, thus "player1" is a pointer to the first element. "current_player" is an int and you try to assign a pointer to it. Additionally you treat "current_player" as a pointer to char in several other parts of your code. You should therefore declare it as a pointer to char.

Code:

printf("\n\n\ About: This is a fun game between two players which will challenge\n\tyour logic and strategy");

The third backslash ('\') isn't necessary.

Code:

scanf("%s",&player1);

As said above, "player1" is a pointer thus you don't need the &-operator.

Code:

if(num>2)

What if the player enters a number below 1? If s/he enters a negative number the balloon count rises.

Code:

for ( j = 1; j <= x; ++j )

In this particular case it's just a matter of style but a more idiomatic way to write the loop is:

Code:

for (j = 0; j < x; ++j)

In some cases (especially in regard to arrays) comparisons with "<=" in a loop could introduce subtle bugs.

Code:

fflush(stdin);
printf("");
scanf("%s",dummy);

http://faq.cprogramming.com/cgi-bin/...&id=1043284351
And what should the last two lines do?

Bye, Andreas

Originally Posted by AndiPersti

Have you already tried to compile your code with all warnings on?
You'll get a bunch of warnings mostly because of this:

Code:

char player1[20],player2[20];
int current_player=player1;

"player1" (and "player2") is an array of char, thus "player1" is a pointer to the first element. "current_player" is an int and you try to assign a pointer to it. Additionally you treat "current_player" as a pointer to char in several other parts of your code. You should therefore declare it as a pointer to char.

Code:

printf("\n\n\ About: This is a fun game between two players which will challenge\n\tyour logic and strategy");

The third backslash ('\') isn't necessary.

Code:

scanf("%s",&player1);

As said above, "player1" is a pointer thus you don't need the &-operator.

Code:

if(num>2)

What if the player enters a number below 1? If s/he enters a negative number the balloon count rises.

Code:

for ( j = 1; j <= x; ++j )

In this particular case it's just a matter of style but a more idiomatic way to write the loop is:

Code:

for (j = 0; j < x; ++j)

In some cases (especially in regard to arrays) comparisons with "<=" in a loop could introduce subtle bugs.

Code:

fflush(stdin);
printf("");
scanf("%s",dummy);

FAQ > Why fflush(stdin) is wrong - Cprogramming.com
And what should the last two lines do?

Bye, Andreas

Thank you for pointing out all those junk

You're right about the warnings, and you're right about the incorrect logic too!!

The last two lines do nothing..they just keep the program open, maybe i should do it some other way

Got 96/100 just because when 1 balloon remains and the player enters 2, it says "-1 balloons remaining"

I don't know why I missed such simple logic, well, for now I've got some assignments and tests, will post a revised solution very soon!!

Thanks once again for taking the time to read my code.