2d array relative to direction on grid - c++

Question

I'm working on a program that hides a number in a grid (2d array, 10 columns and 10 rows labeled with array filled with a character such as a . or -) and prints the grid to screen.
a num for row and column are random generated and saved to array as an 'X', only shown when found
The user enters the row and column....then displays that user's guess on the grid to the screen as a character. after that, choose which direction to move...and give hints on where the randomized location for that item is hidden in the array within ten tries total or less(user cannot go to previous guess played during the program).
for example:
while (guess <9 || found == false){
cout << "Enter your next move....n, e, s, w, ne, nw, se, sw";
cin >> d;
if (d == 'E' || d == 'e')
{
col++;
guess++;
array[col][row] = guess;
cout << "moved east";
}
else if (d == 'W' || d == 'w')
{
col--;
guess++;
array[col][row] = guess;
cout << "west...";
}
....etc
else {
guess++;
cout << "not valid move. turn is incremented";
}
then after the entire while loop on all those, i display the array again but I can't get the moves to show up other than the first step(need it to do 1, 2, 3...10 or until found=true to display 'X') and am not sure how to give the hint on where the item is hidden based on guess direction. if the item is north above where the user guessed their first starting place(column and row), it should display hint based on that location. How could I implement that?

Answer 1

Must be: while(guess < 9&&!found), either condition not met shall leave to exiting, so you only stay in the loop if both are met... Side note: It is better code style not to compare booleans, just use if(condition) or if(!condition) as in the fixed loop condition.
Be aware that you should check your board's borders:if(col > 0)
{
--col;
++guess;Analogously for the other directions...
You need to check, if the field intended to visit already has been visited. Assuming the fields unvisited carry value 0, you can add the following condition to the border check:if(col > 0 && array[col - 1][row] != 0)Again, other directions analogously.
For the hints, calculate the distances of the coordinates between target and current position; if negative, add 'n' or 'w' to your hint for vertical or horizontal direction respectively, if positive, 's' or 'e', otherwise, no hint for the respective direction (easiest variant, prefers diagonal directions first).
Side note array[col][row]: Typically, you name the indices of the array the other way round: array[row][col]; if iterating, you use outer loop for rows, inner loop for cols (to profit from data locality):
for(r = 0; r < rows; ++r)
for(c = 0; c < cols; ++c)
std::cout << getCharFor(array[r][c]);
North and south, with code above, still are associated with vertical, east and west with horizontal direction...
Edit in response to your comment:
You do not need to check the border you are moving away from...if(d == "e")
{
if(/*col > 0 &&*/ col <= 10)
//^^^^^^^^^ obsolete...
}
If your array has a size of NxN, then the maximum index you can access is N - 1, so you only can go east if you are currently maximally at position N - 2 (if(col < N-1))
If you go diagonally, you have to check both horizontal and vertical direction:if(d == "sw")
{
if(row < N - 1 && col > 0)
}
If you have determined that you do not exit the board with the designated move, then you can access the field in the designated direction:if(d == "sw")
{
if(array[row + 1][col - 1] != 0)
// visited already; assuming you have initialized the array with 0...
}
Side note: For the hints, there is yet an unconsidered problem left: What, if the user does not follow them, and due to this, the program gives a hint to a field that already has been visited???