About Tic-Toc-Toe Game in C language - c

Question

I wrote a full code for this game using do and while in C language .. but I stop in point which I don't know how I can type a code to decide if the game is drawn or not..
Thanks
This is my code
#include <stdio.h>
#include <stdlib.h>
char matrix[3][3];
char check(void);
void init_matrix(void);
void get_player1_move(void);
void get_player2_move(void);
void disp_matrix(void);
int main(void)
{
char done,choise;
printf("Welcome to the tic-tac-toe game!!!\n\n");
printf("Rule for playing the game is:\n\n");
printf("Each player must put the value of raw and column like: 1 2 to put his symbol in\nthe tic-tac-toe board.\n\n");
printf("The tic-tac-toe board looks like as follows:\n\n");
init_matrix();
disp_matrix();
printf("Are you ready to start the game? ");
scanf(" %c",&choise);
if (choise == 'y'){
do {
get_player1_move();
done = check(); /* if winner or not */
if(done!= ' ') break; /* winner!*/
disp_matrix();
get_player2_move();
disp_matrix();
done = check(); /* if winner or not */
if(done!= ' ') break; /* winner!*/
} while(done== ' ');
if(done=='X') printf("Player 1 won the game!!!!!\n");
else printf("Player 2 won the game!!!!!\n");
}
else {
printf("\n\nThank you!!!\n\n");
printf("We hope you will play the game anther time....");
}
return 0;
}
/****************************************************/
void init_matrix(void)
{
int i, j;
for(i=0; i<3; i++)
for(j=0; j<3; j++) matrix[i][j] = ' ';
}
/****************************************************/
void get_player1_move(void)
{
int x, y;
printf("Enter row and column input for player 1: ");
scanf("%d%*c%d", &x, &y);
x--; y--;
if(matrix[x][y]!= ' '){
printf("You can not choose this row and clumn!! Try again\n");
get_player1_move();
}
else matrix[x][y] = '1';
}
/****************************************************/
void get_player2_move(void)
{
int x, y;
printf("Enter row and column input for player 2: ");
scanf("%d%*c%d", &x, &y);
x--; y--;
if(matrix[x][y]!= ' '){
printf("You can not choose this row and clumn!! Try again\n");
get_player2_move();
}
else matrix[x][y] = '2';
}
/****************************************************/
void disp_matrix(void)
{
int t;
for(t=0; t<3; t++) {
printf(" : : ");
printf("\n %c : %c : %c",matrix[t][0],matrix[t][1], matrix [t][2]);
if(t!=2) printf("\n ----:----:----\n");
}
printf("\n\n");
}
/****************************************************/
char check(void)
{
int i;
for(i=0; i<3; i++) /* check rows */
if(matrix[i][0]==matrix[i][1] && matrix[i][0]==matrix[i][2])
return matrix[i][0];
for(i=0; i<3; i++) /* check columns */
if(matrix[0][i]==matrix[1][i] && matrix[0][i]==matrix[2][i])
return matrix[0][i];
/* test diagonals */
if(matrix[0][0]==matrix[1][1] && matrix[1][1]==matrix[2][2])
return matrix[0][0];
if(matrix[0][2]==matrix[1][1] && matrix[1][1]==matrix[2][0])
return matrix[0][2];
return ' ';
}

Answer 1

if(done=='X') should be if(done == '1') because matrix contains 1 and 2, not X and O.
To tell if the game is a draw, keep a count of the number of moves, and break out of the loop when it reaches 9. You only have to check this after player 1 moves, because player 2 always moves on even numbers.
There's no need to use do{...} while (done == ' ') because the code always breaks out of the loop when done != ' '. So just use while(1) to make an infinite loop.
int count = 0;
while (1) {
get_player1_move();
done = check(); /* if winner or not */
disp_matrix();
if(done!= ' ' || ++count == 9) break; /* winner or draw */
get_player2_move();
disp_matrix();
done = check(); /* if winner or not */
if(done!= ' ') break; /* winner!*/
++count;
}
if (done == ' ') {
printf("It's a draw!\n");
} else {
printf("Player %c won the game!!\n", done);
}

Answer 2

Set a boolean variable "draw" as false and if no player wins and all 9 moves are done and no free block is available set the "draw" as true. And calculate the result based on "draw" that either the game is a draw or not.