Author eryksun
Recipients Yugi, eryksun, xtreak
Date 2019-08-24.17:31:18
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As Karthikeyan noted, in a regular string literal, backslash is an escape character that's used in the following escape sequences:

    \N{name}   : named character
    \UXXXXXXXX : 32-bit hexadecimal ordinal (e.g. \U0010ffff)
    \uXXXX     : 16-bit hexadecimal ordinal (e.g. \uffff)
    \xXX       : 8-bit hexadecimal ordinal (e.g. \xff)
    \OOO       : 9-bit octal ordinal (e.g. \777)
    \OO        : 6-bit octal ordinal (e.g. \77)
    \O         : 3-bit octal ordinal (e.g. \7)

    \a : \x07, \N{BEL}, \N{ALERT}
    \b : \x08, \N{BS}, \N{BACKSPACE}
    \n : \x0a, \N{LF}, \N{NL}, \N{LINE FEED}, \N{NEW LINE}
    \f : \x0c, \N{FF}, \N{FORM FEED}
    \r : \x0d, \N{CR}, \N{CARRIAGE RETURN}
    \" : \x22, \N{QUOTATION MARK}
    \' : \x27, \N{APOSTROPHE}
    \\ : \x5c, \N{REVERSE SOLIDUS}

For a Windows path, either we can use a normal string literal with backslash path separators escaped by doubling them or we can use a raw string literal. 

One corner case with a raw string literal is that it can't end with an odd number of backslashes. We can address this in one of two ways. Either rely on the compiler's implicit concatenation of string literals, or rely on the system's path normalization to collapse multiple path separators (except at the beginning of a path). For example:

    >>> print(r'C:\Users' '\\')
    >>> print(r'C:\Users\\')

The system normalizes the second case to collapse repeated backslashes. For example:

    >>> print(os.path._getfullpathname(r'C:\Users\\'))
    >>> os.path.samefile(r'C:\Users\\', r'C:\Users' '\\')

We can also use forward slash as the path separator for file-system paths (but not registry paths), such as paths that we're passing to open() or os functions. I don't recommend this if a file-system path is to be passed as a command-line argument. Some programs use forward slash as a switch for command-line options. In this case first normalize the path via os.path.normpath, or via replace('/', '\\').

In some cases a path may be returned to us in Windows with a "\\?\" prefix (backslash only), which is sometimes referred to as an extended path. (More specifically, it's a native path in the device namespace.) This tells the Windows API to skip path normalization. If a path begins with exactly this prefix, then appending components to it with forward slash results in a path that will not work. Use os.path.join, or normalize the path via os.path.normpath to ensure the final path uses only backslash as the path separator.
Date User Action Args
2019-08-24 17:31:18eryksunsetrecipients: + eryksun, xtreak, Yugi
2019-08-24 17:31:18eryksunsetmessageid: <>
2019-08-24 17:31:18eryksunlinkissue37939 messages
2019-08-24 17:31:18eryksuncreate