Create A Game of Tic Tac Toe in C Language Assignment Solutions

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

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; }