thats what i am talking about, there will be a problem when you check the neighbours as it stands.
You can step through the board in a nested loop of two yes. But then you need to check each cell to find out if it lives or dies on the next generation. It should be clear to you that each cell is always thought of as being surrounded by eight neighbours that 'touch it' so a further loop to examine the immediate neighbours is required. There are other fancy ways to optimally examine the board but in your case stick with the method discussed thus far
Now given that some cells are at the edges of the board, what do you think the implications of this are?
C = current cell,
For example imagine it is the top left cell of your board as you presently have it.