Tic-Tac-Toe program in about 250 lines

Boolean values

C99 has Boolean values: #include <stdbool.h>.

Variable names

int x=0; int j=0; 

x and j don't describe what they do. Is j the turn number? Also, I don't see x used anyway except this commented out line: //turn[j] = x;.

Repetition

I would set up a variable name to reduce all those ifs. Maybe something like:

char token = (j % 2 == 0) ? 'O' : 'X'; if (turn[j]==1) { board[0][0] = token; printf("\n\n"); printf(" %c | %c | %c\n", board[0][0], board[0][1], board[0][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[1][0], board[1][1], board[1][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[2][0], board[2][1], board[2][2]); } // ... 

Also, nested for loops may be useful here to reduce that list of ifs to one value.

Edit:

In clarification to the comments, the ?: operator is a ternary conditional operator. It works as follows:

condition ? (run this statement if condition evaluated to true) : (run this statement if condition evaluated to false); 

In this example, I used it in combination with an assignment operation.

C99 is informal terminology to mean ISO/IEC 9899:1999, which was finalized in 1999. The latest version of C is C11, which was finalized in 2011, and which should also support Boolean variables.

From a quick glance, this snippet looks like it could be extracted into its own parameterized function:

 printf("\n\n"); printf(" %c | %c | %c\n", board[0][0], board[0][1], board[0][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[1][0], board[1][1], board[1][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[2][0], board[2][1], board[2][2]); 

DRY - Don't Repeat Yourself ;-)

Notes:

• It seems that you don't need to store the input in the turn array
• You are not asserting legal user input (values between 1 and 9)
• You are not asserting that the user inputs a new value at every turn
• Since user input is a single character, you may as well scan for "%c"
• You can calculate the indexes to board based on the user input (turn)
• You can calculate the player (x or o) based on the iteration number (j)
• You are not changing the value of winner (initialized as false) anywhere

Implementation:

int j; char turn; int index; char players[] = {'o','x'}; for (j=0; j<9; j++) { while (1) { scanf("%c",&turn); if (turn == '\n') continue; // skip the newline character index = turn-'1'; if (!(0 <= index && index <= 8)) printf("Invalid input\n"); else if (board[index/3][index%3] == players[0] || board[index/3][index%3] == players[1]) printf("Already chosen\n"); else break; } board[index/3][index%3] = players[j%2]; printf("\n\n"); printf(" %c | %c | %c\n", board[0][0], board[0][1], board[0][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[1][0], board[1][1], board[1][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[2][0], board[2][1], board[2][2]); /* if there is a winner { print the winner break; } */ } 

As you can see, I left it for you to check if there is a winner...

Some big problems in your code others haven't mentioned:

1. You use strcasecmp. This function is non-standard. This makes your code non-portable.

2. You use

   scanf("%s", scanned); 

   and scanf will happily scan unlimited characters and write it into scanned. You will (and already does) suffer from a buffer overflow. scanned can hold a maximum of 2 characters (+1 for the NUL-terminator). Typing more than two characters leads to Undefined Behavior as other characters are written into an invalid memory location.

   Fix this problem by changing

   char scanned[3]; 

   to

   char scanned[4]; /*Can hold a max of 3 chars (+1 for the NUL-terminator at the end) 

   and

   scanf("%s", scanned); 

   to

   scanf("%3s", scanned); /*Scan a max of 3 chars and append a NUL-terminator at the end*/ 

   Do note that you might have to clear the input buffer to remove extra characters (if the user inputs more than 3 characters).

3. The function playGame() is designed to return an int. If the user types in "yes" for the scanf in playGame(), the if will execute and the function exits without a return statement. This leads to Undefined Behavior. Fix it by adding return 0; at the end of that function.

4. The function startGame() is also designed to return an int, but you never return an int from it. This, again, leads to Undefined Behavior. If you do not want it to return a value change

   int startGame() 

   to

   void startGame() 

   which indicates that the function will not return any value.

Other issues:

• You have a lot of repeating code:

  printf("\n\n"); printf(" %c | %c | %c\n", board[0][0], board[0][1], board[0][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[1][0], board[1][1], board[1][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[2][0], board[2][1], board[2][2]); 

  which can be put into a function named printBoard():

  void printBoard() { printf("\n\n"); printf(" %c | %c | %c\n", board[0][0], board[0][1], board[0][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[1][0], board[1][1], board[1][2]); printf("-----------\n"); printf(" %c | %c | %c\n", board[2][0], board[2][1], board[2][2]); } 

  The logic used can also be simplified by using loops:

  void printBoard() { printf("\n\n"); for(int i = 0; i < 3; i++) { printf(" %c | %c | %c\n", board[i][0], board[i][1], board[i][2]); printf("-----------\n"); } } 

  But this also prints the last printf an extra time, so use

  void printBoard() { printf("\n\n"); for(int i = 0; i < 3; i++) { printf(" %c | %c | %c\n", board[i][0], board[i][1], board[i][2]); if(i != 2) { printf("-----------\n"); } } } 

• Here:

  scanf("%d",&turn[j]); 

  You don't check if user enters a valid number. You also store the inputs in an array. Why? So this:

  int turn[9]; 

  can be

  int turn; 

  Checking the return value of scanf will reveal if the user enters a number or something else. If your scanf returns 1, it indicates that scanf successfully got a number as input and if it returns 0, it indicates that scanf failed to scan a number. It can also return -1(EOF) if it encounters a special value called EOF.

• Use meaningful variable names instead of x,scanned etc. I leave this up to you.

• This:

  else { if (strcmp(scanned,"no")==0 || strcmp(scanned,"nah")==0 || strcmp(scanned,"naw")==0) { printf("That's too bad!/nThis program will now end."); return 1; } printf("Not valid input!/nThis program will now end."); return 1; } 

  can be written as

  else { char* exit_strings[] = {"no", "nah", "naw"}; for(int i = 0; i < sizeof(exit_strings); i++) if (strcmp(scanned, exit_strings[i]) == 0) { printf("That's too bad!\n"); printf("This program will now end.\n"); return 1; } printf("Not valid input!\n"); printf("This program will now end.\n"); return 1; } 

  making your code better. Note that you used /n while \n indicates a newline character.

Here is a revised working version of your code.

It includes the usage of boolean variables; deals with code-repetition; implements cases of wrong user inputs; and decides stalemate, winner cases.

#include <stdio.h> #include <string.h> void print(char board[][3]) { for (int i=0; i<3; i++) { printf(" %c | %c | %c\n", board[i][0], board[i][1], board[i][2]); if(i!=2) printf("-----------\n"); } } bool winnerCheck(char board[][3], char check) { bool winner = board[0][0]==check && board[0][1]==check && board[0][2]==check || board[1][0]==check && board[1][1]==check && board[1][2]==check || board[2][0]==check && board[2][1]==check && board[2][2]==check || board[0][0]==check && board[1][0]==check && board[2][0]==check || board[0][1]==check && board[1][1]==check && board[2][1]==check || board[0][2]==check && board[1][2]==check && board[2][2]==check || board[0][0]==check && board[1][1]==check && board[2][2]==check || board[0][2]==check && board[1][1]==check && board[2][0]==check; if (winner) printf("Player %d wins!\n", (check=='O')?1:2); return winner; } void startGame() { //Sets up board for values 1-9; char board[3][3] = { {'1','2','3'}, {'4','5','6'}, {'7','8','9'} }; bool winner = false; printf("\nHere is your playing board. Player One is Os and Player Two is Xs\n"); printf("Entering a number 1-9 (then pushing enter) as shown below will use\nthe current Player's turn in that location.\n\n"); print(board); int j=0,cnt=9,turn; while (j<cnt && winner==false) { scanf("%d",&turn); if(board[turn/3][turn%3-1]!='O' && board[turn/3][turn%3-1]!='X') board[turn/3][turn%3-1] = (j%2==0)?'O':'X'; else { printf("Illegal move\n"); continue; } printf("\n\n"); print(board); j++; winner = winnerCheck(board,'O') || winnerCheck(board,'X'); } if(!winner) printf("Stalemate!\n"); } int playGame() { char scanned[10]; printf("Do you wish to play tick-tack-toe?\n"); scanf("%s", scanned); if(strcasecmp(scanned,"yes")==0) startGame(); else { if (strcmp(scanned,"no")==0 || strcmp(scanned,"nah")==0 || strcmp(scanned,"naw")==0) printf("That's too bad!\nThis program will now end."); else printf("Not valid input!\nThis program will now end."); return 1; } } int main() { playGame(); return 0; } 

The changes are listed out:

• One of the reasons to tackle the increasing length of your code was board printing repetition of code. This is handled by the usage of print(char board[][3]) function

• Your code would not stop in case one of the winners has won; and at the end of the program, in case none of the users win, the program simply terminates. The winnerCheck(char board[][3], char check) takes in the board configuration and checks for the first/second user's winning condition. And after detecting a successful triumph, terminates after a notification to the players.

• In case a user tries to make an illegal move at a location already occupied, the line of code : if(board[turn/3][turn%3-1]!='O' && board[turn/3][turn%3-1]!='X') takes care of this situation.

• In case there is no winner, the program checks for the winner variable and notifies the players.

• The else part of code in the playgame() function has repeated return 1; statements. The revised working code addresses the issue.

• The turn array is not necessary and rather a character turn variable is used for taking in the user inputs. One may check the validity of the user's input as well since the number should be in [1-9] range. The board coordinates to be set can be decided using (turn/3,turn%3-1) rather than a set of if statements.

• Something you may try: The user may enter a word longer than 3 characters; thus use a character array of longer than 3 length, say 10 charaters. But the user may use a 11 character word. Hence, as a solution, one may try to address this by specifying a boolean choice of [0/1] in the terminal and prompt the user to enter again if the choice is incorrect, since it may be possible that the user enters affirmative or negation words in different language or dialect.

My answer is to use a "switch/case" expression. Here's how it works:

switch (expression that evaluates to int) { case 1: { // code to be executed if the expression evaluates to 1 break; } // add as many cases as needed default: { // this is the "else" case. break; } } 

This way, you won't need to write statements such as these:

if (turn[j] == 1) { // code } 

You can replace them with this:

switch(turn[j]) { case 1: { // code break; } case 2: { // more code break; } default { // it equals neither 1 nor 2 break; } } 

In addition, instead of explicitly stating conditions for "if"s can only run if something is 0, you can simply remove the "==0" and put a binary "not" operator in front of the condition. (0 is binary for "false")

When you start the game, you can remove all of the "if" statements within what should be if (!(j%2)) and replace it with this:

board[(turn[j] - 1) / 3][(turn[j] - 1) % 3] = '0' printf("\n\n %c | %c | %c\n-----------\n %c | %c | %c\n-----------\n %c | %c | %c\n", board[0][0], board[0][1], board[0][2], board[1][0], board[1][1], board[1][2], board[2][0], board[2][1], board[2][2]);