Shoutbox

Python code:

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import pyglet as pyg
import random
from pyglet.window import key
window=pyg.window.Window()
 
#map- when arrow keys are pressed in pyglet, the symbol that represents them is a number
#he numbers are:
#65361 = left = 0
#65362 = up = 1
#65363 = right = 2
#65364 = down = 3
#we now have 4 numbers 1-4, each representing an arrow key
#
keys=[False]*4
#so now we have a list of 4 False values, which will represent each arrow,
#python indexes an array/list starting from 0
 
#above we mention 4 numbers, 65361,65362,65363,65364
#instead of mapping each to a dictionary if we get the last digit and subtract 1 we end up with a set like
#0,1,2,3
#so now if we let this number index keys like this
#keys[((int(str(symbol)[-1]))-1)]
#it is the same as saying:
#
#def on_key_press(symbol,modifiers,keys=keys):
#   if symbol == 65361:
#       keys[0]=True    #keys[0] is left
#   if symbol == 65362:
#       keys[1] == True
# #ect,ect
 
#keys are mapped ((int(str(symbol)[-1]))-1)
 
@window.event
def on_key_press(symbol,modifiers,keys=keys):
    #print symbol,((int(str(symbol)[-1]))-1)
    keys[((int(str(symbol)[-1]))-1)] = True
@window.event
 
def on_key_release(symbol,modifiers,keys=keys):
    keys[((int(str(symbol)[-1]))-1)] = False
 
def update(dt,keys=keys):
    #by using this statement we determine more complex events first (left + up) so it fulfils these instead fulfilling a less complex (left)
    if keys[0] and keys[1]:
        print 'left'+'up'
    elif keys[1] and keys[2]:
        print 'right'+'up'
    elif keys[0] and keys[3]:
        print 'down'+'left'
    elif keys[2] and keys[3]:
        print "down"+'right'
    elif keys[0]:
        print "left"    #due to this, if two conflicting keys are help the upper most one will occur
    elif keys[1]:
        print "up"      #up occurs if 'down' and 'up' are down, left occurs if 'left' and 'right' are down
    elif keys[2]:
        print 'right'
    elif keys[3]:
        print "down"
   
 
pyg.clock.schedule_interval(update, 0.1)
pyg.app.run()

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