Question

What is the correct syntax of open() function?

(a) file = open(file_name [, access_mode][, buffering])

(b) file object = open(file_name [, access_mode][, buffering])

(c) file object = open(file_name)

(d) none of the mentioned

I have been asked this question in an online interview.

Enquiry is from Files in division Regular Expressions and Files of Python

Answer 1

Correct option is (b) file object = open(file_name [, access_mode][, buffering])

To explain I would say: Open() function correct syntax with the parameter details as shown below:

file object = open(file_name [, access_mode][, buffering])

Here is parameters’ detail:

file_name: The file_name argument is a string value that contains the name of the file that you want to access.

access_mode: The access_mode determines the mode in which the file has to be opened, i.e., read, write, append, etc. A complete list of possible values is given below in the table. This is optional parameter and the default file access mode is read (r).

buffering: If the buffering value is set to 0, no buffering will take place. If the buffering value is 1, line buffering will be performed while accessing a file. If you specify the buffering value as an integer greater than 1, then buffering action will be performed with the indicated buffer size. If negative, the buffer size is the system default(default behavior).