python中函数的可选参数

Question

I'm trying to write slicing code which gets a linked list, start,stop, step. My code is supposed to behave like same as using list[start:step:stop] .

My problem starts when the user insert only 1 argument (let's says it was x ) - then, x is supposed to get into stop and and not start . But, I have been told that optional arguments need to appear at the end of all the arguments.

Can someone tell me how can we insert only 1 input to the second argument while the first one is mandatory but the second one is not? By the way, i can't use built-in functions

Answer 1

You could write a LinkedList class that defines the __ getitem __ function to get access to python's notation.

class LinkedList:

    # Implement the Linked ...

    def __getitem__(self, slice):

        start = slice.start
        stop = slice.stop
        step = slice.step

        # Implement the function

Then you can use LinkedList like you want

l = LinkedList()
l[1]
l[1:10]
l[1:10:2]

Answer 2

You could try:

def myF(*args):
    number_args = len(args)
    if number_args == 1:
        stop = ...
    elif number_args == 2:
        ...
    elif number_args == 3:
        ...
    else
        print "Error"

*args means that the arguments passed to the function myF will be stored in the variable args .

Answer 3

Using optional (named) arguments:

def foo(start, stop=None, step=1):
    if stop == None:
        start, stop = 0, start
    #rest of the code goes here

Then foo(5) == foo(0,5,1) , but foo(1,5) == foo(1,5,1) . I think this works, anyway... :)

Answer 4

def func(arg1=None, arg2=None, arg3=None):
    if not arg1:
        print 'arg1 is mandatory'
        return False
    ## do your stuff here

arg1 = 1
arg2 = 4
arg3 = 1

## works fine
func(arg1=arg1, arg2=arg2, arg3=arg3)

## Evaluated as False, Since the first argument is missing
func(arg2=arg2, arg3=arg3)