Python – How to invoke the super constructor in Python

classconstructorinheritancepythonsuperclass

class A:
    def __init__(self):
        print("world")

class B(A):
    def __init__(self):
       print("hello")

B()  # output: hello

In all other languages I've worked with the super constructor is invoked implicitly. How does one invoke it in Python? I would expect super(self) but this doesn't work.

Best Answer

In line with the other answers, there are multiple ways to call super class methods (including the constructor), however in Python-3.x the process has been simplified:

Python-2.x

class A(object):
 def __init__(self):
   print "world"

class B(A):
 def __init__(self):
   print "hello"
   super(B, self).__init__()

Python-3.x

class A(object):
 def __init__(self):
   print("world")

class B(A):
 def __init__(self):
   print("hello")
   super().__init__()

super() is now equivalent to super(<containing classname>, self) as per the docs.

Related Solutions

C# – Calling the base constructor in C#

Modify your constructor to the following so that it calls the base class constructor properly:

public class MyExceptionClass : Exception
{
    public MyExceptionClass(string message, string extrainfo) : base(message)
    {
        //other stuff here
    }
}

Note that a constructor is not something that you can call anytime within a method. That's the reason you're getting errors in your call in the constructor body.

Python – How to check whether a file exists without exceptions

If the reason you're checking is so you can do something like if file_exists: open_it(), it's safer to use a try around the attempt to open it. Checking and then opening risks the file being deleted or moved or something between when you check and when you try to open it.

If you're not planning to open the file immediately, you can use os.path.isfile

Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.

import os.path
os.path.isfile(fname)

if you need to be sure it's a file.

Starting with Python 3.4, the pathlib module offers an object-oriented approach (backported to pathlib2 in Python 2.7):

from pathlib import Path

my_file = Path("/path/to/file")
if my_file.is_file():
    # file exists

To check a directory, do:

if my_file.is_dir():
    # directory exists

To check whether a Path object exists independently of whether is it a file or directory, use exists():

if my_file.exists():
    # path exists

You can also use resolve(strict=True) in a try block:

try:
    my_abs_path = my_file.resolve(strict=True)
except FileNotFoundError:
    # doesn't exist
else:
    # exists

Best Answer

Related Solutions

C# – Calling the base constructor in C#

Python – How to check whether a file exists without exceptions

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