Eryk Sun <[email protected]> added the comment:
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}
\t : \x09, \N{HT}, \N{TAB}, \N{CHARACTER TABULATION}, \N{HORIZONTAL
TABULATION}
\n : \x0a, \N{LF}, \N{NL}, \N{LINE FEED}, \N{NEW LINE}
\v : \x0b, \N{VT}, \N{LINE TABULATION}, \N{VERTICAL TABULATION}
\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' '\\')
C:\Users\
>>> print(r'C:\Users\\')
C:\Users\\
The system normalizes the second case to collapse repeated backslashes. For
example:
>>> print(os.path._getfullpathname(r'C:\Users\\'))
C:\Users\
>>> os.path.samefile(r'C:\Users\\', r'C:\Users' '\\')
True
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.
----------
nosy: +eryksun
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<https://bugs.python.org/issue37939>
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