[CC][Unfinished] - Coord Password Idea

--[[
    Unique password concept
    Idea and created by VintageRIce
    Started: July 26th 2014
]]

--Setting up the variables
local xPass = {"1", "2", "3", "4"}
local yPass = {"1", "2", "3", "4"}
local passChars = {"A", "B", "C", "D"}

local xLoc = {0, 0, 0, 0}
local yLoc = {0, 0, 0, 0}

local xPos = 1
local yPos = 1

local doCount = 1
local passCount = 0

--Check if the password is correct
function checkPass()
    for i = 1, 4 do
        if xLoc[i] == xPass[i] then
            passCount = passCount + 1
        end

        if yLoc[i] == yPass[i] then
            passCount = passCount + 1
        end
    end

    term.clear()
    term.setCursorPos(1, 1)
    print(passCount)
end

--Prints the GUI on each key press
function printScreen(value)
    term.clear()

    for i = 1, 51 do
        paintutils.drawPixel(i, 16, colours.red)
    end

    term.setCursorPos(1, 17)
    print("Editing: "..passChars[doCount])
    print("Current Position: "..xPos..", "..yPos)

    term.setCursorPos(xPos, yPos)
    print("X")
end

--Monitors the key input
function monitorInput()
    while true do
        event, char = os.pullEvent("char")

        if char == "w" or "a" or "s" or "d" or "f" then
            if char == "w" then
                yPos = yPos - 1
            end

            if char == "a" then
                xPos = xPos - 1
            end

            if char == "s" then
                yPos = yPos + 1
            end

            if char == "d" then
                xPos = xPos + 1
            end

            if char == "f" then
                for i = 1, 4 do
                    if doCount == i then
                        xLoc[i] = xPos
                        yLoc[i] = yPos
                    end
                end

                doCount = doCount + 1
            end

            printScreen(doCount)
        end
    end
end

printScreen()
monitorInput()