__init__.py 文件源码

def debugmsg(msg, msgTupleLambda=None):
        '''
            debugmsg - If #IS_DEVELOPER_DEBUG is changed to True (in the code, changing it after import has no affect ) this will print a debug message

                        to stderr prefixed with "DEBUG: " and suffixed with a newline.

                       Don't create your message like:

                           debugmsg( "Order(id=%s) has %d pepperoni pizzas" %( myOrder.id, len( [ topping for pizza in getPizzas() for topping in pizza.toppings() if 'pepperoni' in topping ] ) ) )

                       Which would evaluate that comprehension whether or not debug mode is on,

                         Instead, use a lambda to return that same tuple, so that code is only evaluated in the #debugmsg function calls your lambda:

                           debugmsg( "Order(id=%s) has %d pepperoni pizzas", lambda : ( myOrder.id, len( [ topping for pizza in getPizzas() for topping in pizza.toppings() if 'pepperoni' in topping ] ) ) )
        '''

        if issubclass(msgTupleLambda.__class__, (tuple, list)):
            sys.stderr.write('''WARNING: debugmsg called with a tuple/list for msgTupleLambda.\n\tShould be a lambda that returns the tuple, so when IS_DEVELOPER_DEBUG=False it is not evaluated. \n\n''' + ''.join(traceback.format_stack()[:-1]) + "\n\n" )

            msgTuple = msgTupleLambda
        else:
            msgTuple = msgTupleLambda()

        sys.stderr.write('DEBUG: %s\n' % msgTuple )