To Create A Game Of Tic Tac Toe In C -Skeleton Code Provided

Instructions

Objective

Great job on taking on the challenge to create a game of Tic Tac Toe in C language! It's a fantastic opportunity to showcase your programming skills and problem-solving abilities. I'm confident that you can complete a C assignment like this with your dedication and passion for coding. Remember, Tic Tac Toe is a classic game with simple rules, but implementing it in C requires attention to detail and logical thinking. Break down the task into smaller steps, and don't hesitate to refer to C language resources for guidance.

Requirements and Specifications

Skeleton code

#include
#include //time()
#include //srand() and rand()
typedef struct
{
   int occupied; // 0 means unoccupied, 1 means occupied
   char symbol; //’X’ or ’O’ or ’ ’
} Cell;
typedef struct
{
   Cell cells[3][3]; // 9 spaces in our board
   int winner; // 1 if the currentPlayer has won, 0 otherwise
   int counter; // for how many turns have been played in total
   int size; // should match the dimensions of the ‘cells‘ array
   int currentPlayer; // 0 for ’O’s and 1 for ’X’s
} Board;
// In function prototypes, just the type of each argument is sufficient.
Cell initializeCell(char, int);
void getPosition(Board, int[2]);
Board updateBoard(int, int, Board);
int isWinningMove(int, int, Board);
void printBoard(Board);
int main(void)
{
   srand(time(NULL));
   Board mainBoard;
   // TODO: Initialize the members (except for ‘cells‘) of the mainBoard
   // to appropriate values. You must choose a random player to start.
   // TODO: Initialize each cell in the mainBoard to an unoccupied cell
   // with a single space for the symbol. Note that you’ll need a nested
   // for loop to do this, since the ‘cells‘ array inside mainBoard is 2D.
   // Use your initializeCell() function in the body of the nested for
   // loop to do this.
   while ((mainBoard.winner == 0) && (mainBoard.counter < 9))
   {
       // TODO:
       // 1) Alternate the mainBoard.currentPlayer
       // 2) Create an array that holds 2 ints called ‘position‘
       // 3) call the getPosition() function, passing in the ‘mainBoard‘
       // and the ‘position‘ array. getPosition() "returns" two
       // values by using this array to hold the output values.
       // 4) call updateBoard(), passing in the row and column which
       // were stored in the ‘position‘ array by getPosition() and
       // the ‘mainBoard‘. Assign its return value to mainBoard, effectively
       // updating main()’s local mainBoard variable.
       // 5) set ‘mainBoard.winner‘ to the return value of isWinningMove().
       // Pass in the row and column of the cell that was just played
       // and the ‘mainBoard‘.
       // 6) call printBoard()
}
   if (mainBoard.winner==0)
       printf("Tie Game\n");
else
printf("Player %d won\n", mainBoard.currentPlayer);
return 0; }
void printBoard(Board board)
{
   // Don’t touch this function. It already prints things nicely for you.
   // Note that here, I’m using the first index of the cells array
   // as the row, and the second as the column.
   for (int row = 0; row < board.size; row++)
   {
       printf("| ");
       for (int column = 0; column < board.size; column++)
       {
           printf("%c | ", board.cells[row][column].symbol);
       }
       printf("\n");
   }
   printf("\n");
}
Cell initializeCell(char sym, int isOccupied)
{
   // TODO:
   // 1) Create a local variable of type Cell and initialize its
   // members according to the arguments passed into this function.
   // 2) Return the local variable.
   //
   // This function is here mainly to show you that you can return structs
   // from a function.
}
void getPosition(Board board, int pos[2])
{
   int r, c; // Remember, these are local variables to the function
   do
   {
       printf("Enter the row and column: ");
       scanf("%d%d", &r, &c);
   } while (/*insert condition here*/);
   // TODO: Fill in the condition of the while loop above. It should continue
   // to ask the user for a row and column as long as the row and column
   // the user has asked for is out of bounds OR if the cell is already
   // occupied in the board.
   // TODO: Assign the desired row ‘r‘ to pos[0] and the desired column ‘c‘
   // to pos[1]. Here we’re using the ‘pos‘ array to "return" two values
   // back to main(), more specifically the row and column of the cell the user
   // wants to play their symbol in.
}
Board updateBoard(int r, int c, Board board)
{
   // TODO:
   // 1) Set the cell associated with the row ‘r‘ and column ‘c‘ to
   // occupied.
   // 2) Incremenet the board counter, since a move has just been made.
   // 3) Set the symbol of the cell associated with ‘r‘ and ‘c‘ to
   // the symbol of the currentPlayer (let player 0 be ’O’s and
   // player ‘ be ’X’s.
       // 4) Return the board.
int isWinningMove(int r, int c, Board board)
{
   int possibleWin = 0;
   // TODO: Create a char variable called ‘playerSymbol‘ and set
   // it to either ’X’ or ’O’ depending on the current player.
   // The code below checks to see if there is a 3-in-a-row in the
   // row that was just played in (‘r‘ is the row that was just played in).
   // Do not change this. It is provided to help you with the other
   // checks below. For full credit, you must use a for loop (like below)
   // for each check.
   for (int column = 0; column < board.size; column++)
   {
       if (board.cells[r][column].symbol == playerSymbol)
           possibleWin = 1;
else
       {
           possibleWin = 0;
break; }
}
   if (possibleWin == 1)
   {
       // possibleWin can only be a 1 if all 3 symbols in the row matched
       // so we have a winner!
       return possibleWin;
}
   // TODO: Using the above example on how to check for a 3-in-a-row in
   // the row that was just played in, check for a 3-in-a-row in the
   // column that was just played in. Again, you should do this using
   // a single for loop, and after it, returning ‘possibleWin‘ if
   // ‘possibleWin‘ is a 1.
   // Now we have to check for diagonal 3-in-a-rows, but only if the
   // most recent move was in a diagonal. Let’s check the main diagonal
   // first (top left to bottom right). We can tell if the most recent move
   // is part of the main diagonal if the row and column of the most recent
   // move are identical.
   if (c == r)
   {
       // TODO: Check for a 3-in-a-row in the main diagonal of the board.
       // Use the example for checking for a 3-in-a-row in the row that
       // was just played in.
}
   // Now we check for the off-diagonal (bottom left to top right).
   if (c == board.size - 1 - r)
   {
       // TODO: Check for a 3-in-a-row in the off diagonal of the board.
}
   // If we haven’t returned a 1 at this point after all the checks, then the
   // current player did not just win, so we return a 0 (indicating no one has
   // won yet).
   return 0;
}
}

Screenshots of output

game of Tic Tac Toe in C language

game of Tic Tac Toe in C language 1

Source Code

#include
#include //time()
#include //srand() and rand()
typedef struct
{
    int occupied; // 0 means unoccupied, 1 means occupied
    char symbol; //’X’ or ’O’ or ’ ’
} Cell;
typedef struct
{
    Cell cells[3][3]; // 9 spaces in our board
    int winner; // 1 if the currentPlayer has won, 0 otherwise
    int counter; // for how many turns have been played in total
    int size; // should match the dimensions of the ‘cells‘ array
    int currentPlayer; // 0 for ’O’s and 1 for ’X’s
} Board;
// In function prototypes, just the type of each argument is sufficient.
Cell initializeCell(char, int);
void getPosition(Board, int[2]);
Board updateBoard(int, int, Board);
int isWinningMove(int, int, Board);
void printBoard(Board);
int main(void)
{
    srand(time(NULL));
    Board mainBoard;
    // TODO: Initialize the members (except for ‘cells‘) of the mainBoard
    // to appropriate values. You must choose a random player to start.
    mainBoard.winner = 0;
    mainBoard.counter = 0;
    mainBoard.size = 3;
    mainBoard.currentPlayer = rand() % 2;
    // TODO: Initialize each cell in the mainBoard to an unoccupied cell
    // with a single space for the symbol. Note that you’ll need a nested
    // for loop to do this, since the ‘cells‘ array inside mainBoard is 2D.
    // Use your initializeCell() function in the body of the nested for
    // loop to do this.
    for (int row = 0; row < mainBoard.size; row++)
        for (int col = 0; col < mainBoard.size; col++)
            mainBoard.cells[row][col] = initializeCell(' ', 0);
    while ((mainBoard.winner == 0) && (mainBoard.counter < 9))
    {
        // TODO:
        // 1) Alternate the mainBoard.currentPlayer
        if (mainBoard.currentPlayer == 0)
            mainBoard.currentPlayer = 1;
        else
            mainBoard.currentPlayer = 0;
        // 2) Create an array that holds 2 ints called ‘position‘
        int position[2];
        // 3) call the getPosition() function, passing in the ‘mainBoard‘
        // and the ‘position‘ array. getPosition() "returns" two
        // values by using this array to hold the output values.
        getPosition(mainBoard, position);
        // 4) call updateBoard(), passing in the row and column which
        // were stored in the ‘position‘ array by getPosition() and
        // the ‘mainBoard‘. Assign its return value to mainBoard, effectively
        // updating main()’s local mainBoard variable.
        mainBoard = updateBoard(position[0], position[1], mainBoard);
        // 5) set ‘mainBoard.winner‘ to the return value of isWinningMove().
        // Pass in the row and column of the cell that was just played
        // and the ‘mainBoard‘.
        mainBoard.winner = isWinningMove(position[0], position[1], mainBoard);
        // 6) call printBoard()
        printBoard(mainBoard);
    }
    if (mainBoard.winner==0)
        printf("Tie Game\n");
    else
        printf("Player %d won\n", mainBoard.currentPlayer);
    return 0;
}
void printBoard(Board board)
{
    // Don’t touch this function. It already prints things nicely for you.
    // Note that here, I’m using the first index of the cells array
    // as the row, and the second as the column.
    for (int row = 0; row < board.size; row++)
    {
        printf("| ");
        for (int column = 0; column < board.size; column++)
        {
            printf("%c | ", board.cells[row][column].symbol);
        }
        printf("\n");
    }
    printf("\n");
}
Cell initializeCell(char sym, int isOccupied)
{
    // TODO:
    // 1) Create a local variable of type Cell and initialize its
    // members according to the arguments passed into this function.
    // 2) Return the local variable.
    //
    // This function is here mainly to show you that you can return structs
    // from a function.
    Cell cell;
    cell.occupied = isOccupied;
    cell.symbol = sym;
    return cell;
}
void getPosition(Board board, int pos[2])
{
    int r, c; // Remember, these are local variables to the function
    do
    {
        printf("Enter the row and column: ");
        scanf("%d%d", &r, &c);
    } while (r < 0 || r >= board.size || c < 0 || c >= board.size || board.cells[r][c].occupied);
    // TODO: Fill in the condition of the while loop above. It should continue
    // to ask the user for a row and column as long as the row and column
    // the user has asked for is out of bounds OR if the cell is already
    // occupied in the board.
    // TODO: Assign the desired row ‘r‘ to pos[0] and the desired column ‘c‘
    // to pos[1]. Here we’re using the ‘pos‘ array to "return" two values
    // back to main(), more specifically the row and column of the cell the user
    // wants to play their symbol in.
    pos[0] = r;
    pos[1] = c;
}
Board updateBoard(int r, int c, Board board)
{
    // TODO:
    // 1) Set the cell associated with the row ‘r‘ and column ‘c‘ to
    // occupied.
    board.cells[r][c].occupied = 1;
    // 2) Increment the board counter, since a move has just been made.
    board.counter++;
    // 3) Set the symbol of the cell associated with ‘r‘ and ‘c‘ to
    // the symbol of the currentPlayer (let player 0 be ’O’s and
    // player ‘ be ’X’s.
    if (board.currentPlayer == 0)
        board.cells[r][c].symbol = 'O';
    else
        board.cells[r][c].symbol = 'X';
    // 4) Return the board.
    return board;
}
int isWinningMove(int r, int c, Board board)
{
    int possibleWin = 0;
    // TODO: Create a char variable called ‘playerSymbol‘ and set
    // it to either ’X’ or ’O’ depending on the current player.
    char playerSymbol;
    if (board.currentPlayer == 0)
        playerSymbol = 'O';
    else
        playerSymbol = 'X';
    // The code below checks to see if there is a 3-in-a-row in the
    // row that was just played in (‘r‘ is the row that was just played in).
    // Do not change this. It is provided to help you with the other
    // checks below. For full credit, you must use a for loop (like below)
    // for each check.
    for (int column = 0; column < board.size; column++)
    {
        if (board.cells[r][column].symbol == playerSymbol)
            possibleWin = 1;
        else
        {
            possibleWin = 0;
            break;
        }
    }
    if (possibleWin == 1)
    {
        // possibleWin can only be a 1 if all 3 symbols in the row matched
        // so we have a winner!
        return possibleWin;
    }
    // TODO: Using the above example on how to check for a 3-in-a-row in
    // the row that was just played in, check for a 3-in-a-row in the
    // column that was just played in. Again, you should do this using
    // a single for loop, and after it, returning ‘possibleWin‘ if
    // ‘possibleWin‘ is a 1.
    for (int row = 0; row < board.size; row++)
    {
        if (board.cells[row][c].symbol == playerSymbol)
            possibleWin = 1;
        else
        {
            possibleWin = 0;
            break;
        }
    }
    if (possibleWin == 1)
    {
        return possibleWin;
    }
    // Now we have to check for diagonal 3-in-a-rows, but only if the
    // most recent move was in a diagonal. Let’s check the main diagonal
    // first (top left to bottom right). We can tell if the most recent move
    // is part of the main diagonal if the row and column of the most recent
    // move are identical.
    if (c == r)
    {
        // TODO: Check for a 3-in-a-row in the main diagonal of the board.
        // Use the example for checking for a 3-in-a-row in the row that
        // was just played in.
        for (int row = 0; row < board.size; row++)
        {
            if (board.cells[row][row].symbol == playerSymbol)
                possibleWin = 1;
            else
            {
                possibleWin = 0;
                break;
            }
        }
        if (possibleWin == 1)
        {
            return possibleWin;
        }
    }
    // Now we check for the off-diagonal (bottom left to top right).
    if (c == board.size - 1 - r)
    {
        // TODO: Check for a 3-in-a-row in the off diagonal of the board.
        for (int row = 0; row < board.size; row++)
        {
            if (board.cells[row][board.size - 1 - row].symbol == playerSymbol)
                possibleWin = 1;
            else
            {
                possibleWin = 0;
                break;
            }
        }
        if (possibleWin == 1)
        {
            return possibleWin;
        }
    }
    // If we haven’t returned a 1 at this point after all the checks, then the
    // current player did not just win, so we return a 0 (indicating no one has
    // won yet).
    return 0;
}

Create A Game Of Tic Tac Toe In C Language Assignment Solutions.

Instructions

Requirements and Specifications