CODEWITHSUNDEEP
pythonpointers
alwinlin
alwinlin

Reputation: 723

pointer in python?

When I was trying to figure out the use of imp.load_module in Python, I got the following code (origin page). It's my first time seeing the use of * in Python, is that some pointer like things?

thanks in advance

import imp
import dbgp
info = imp.find_module(modname, dbgp.__path__)
_client = imp.load_module(modname, *info)
sys.modules["_client"] = _client
from _client import *
del sys.modules["_client"], info, _client

Upvotes: 2

Views: 736

Answers (3)

Amit Kumar Rai
Amit Kumar Rai

Reputation: 1

Essentially, they are variables that hold the memory address of another variable.

Upvotes: 0

GWW
GWW

Reputation: 44161

The * on front if info causes the list / tuple to be unwrapped into individual arguments to the function. You can read more about unpacking here in the python documentation. this can also be done with dictionaries for named arguments, see here

For example,

def do_something(a,b,c,d):
    print("{0} {1} {2} {3}".format(a,b,c,d))

a = [1,2,3,4]
do_something(*a)

Output:

1 2 3 4

EDIT:

According to comments to your question by jcomeau_ictx, the operator is called splat

Upvotes: 7

Daenyth
Daenyth

Reputation: 37461

I'm assuming you're talking about the _client = imp.load_module(modname, *info) line.

No, it's not a pointer. It expands the list passed in as an argument. Here's an example:

In [7]: def foo(bar, baz):
   ...:     return bar + baz
   ...: 

In [8]: l = [1,2]

In [9]: foo(l)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)

/home/Daenyth/<ipython console> in <module>()

TypeError: foo() takes exactly 2 arguments (1 given)

In [10]: foo(*l)
Out[10]: 3

There's similar expansion possible for dictionaries.

In [12]: d = {'bar': 1, 'baz': 2}

In [13]: foo(**d)
Out[13]: 3

Upvotes: 4

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