CODEWITHSUNDEEP
pythonfunctiontypes
kiran
kiran

Reputation: 51

How can I specify a pointer as a parameter to a function in Python?

I'm trying to define a function (initdeque()) that takes a pointer to an instance of the deque class. So this is what I tried:

from ctypes import *

class deque(Structure):
        pass

deque._fields_ = [("left",POINTER(node)),("right",POINTER(node))]

def initdeque(class deque *p):
        p.left=p.right=None

But this code gives a syntax error:

  def initdeque(class deque *p):
                      ^
  SyntaxError: invalid syntax

What is the correct syntax?

Upvotes: 3

Views: 212

Answers (2)

John Demetriou
John Demetriou

Reputation: 4381

Python (like Matlab) does not have explicit variable type definition. When a variable is created the compiler defines it's type judging from it's content. So (if I am not wrong) you can call the same function with multiple types as parameters and still work. For example

def fun(p):
       return p=p*5

So u can call fun with a string as a parameter and return you a string that contains 5 times the string you sent. Or you can call it with an integer and the function will return you the result of the multiplication

Upvotes: 0

phihag
phihag

Reputation: 288090

There is no way to specify the type of Python variables, so the declaration of initdequeue should just say:

def initdeque(p):
  p.left = p.right = None

Instead of having a random function doing the initialization, you should consider using an initializer:

class deque(Structure):
  _fields_ = [("left",POINTER(node)), ("right",POINTER(node))]

  def __init__(self):
    self.left = self.right = None

Now, to create a new instance, just write

 p = dequeue()

Upvotes: 6

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