Question

How can I check 3 elements of a single-dimensional array are equal to a specific value?

Background: I am trying to check for the winner of a game of tic-tac-toe (noughts and crosses). In other words check for any lines in a 3×3 grid flattened into a single dimensional array.

Here is the code I have tried:

Attempt 1:

switch ()
{
    case board[(6&3&0) || (7&4&1)]: case board[7&4&1]: case board[8&5&2]: case board[0&1&2]: case board[3&4&5]: case board[6&7&8]: case board[6&4&2]: case board[8&4&0]:
    System.out.println("Equal to x")
    break;
}

Attempt 2:

if (board[7] , board[4] , board[1] == 'X')
system.out.println("Equal");

Attempt 3:

if ( board[6&3&0] == 'X'  || board[7&4&1] == 'X' || board[8&5&2] == 'X' || board[0&1&2] == 'X' || board[3&4&5] == 'X' || board[6&7&8] == 'X' || board[6&4&2] == 'X' || board[8&4&0] == 'X') {
    System.out.println("Equal to x");
}else if
( board[6&3&0] == 'O'  || board[7&4&1] == 'O' || board[8&5&2] == 'O' || board[0&1&2] == 'O' || board[3&4&5] == 'O' || board[6&7&8] == 'X' || board[6&4&2] == 'O' || board[8&4&0] == 'O'){
    System.out.println("Equal to o!");
}

Answer 1

Your “playing-field” for tic-tac-toe looks like this:

0 1 2
3 4 5
6 7 8

With 0-8 being the indices in a char[] array:

marks = new char[9];

To check if we have three ‘x’ in a row:

'x' == marks[0] == marks[3] == marks[6] // a vertical row
'x' == marks[0] == marks[1] == marks[2] // a horizontal row
'x' == marks[0] == marks[4] == marks[8] // a diagonal row

This should get you started.

EDIT: And this will compile:

boolean win = marks[0] == 'x' && marks[3] == 'x' && marks[6] == 'x';