Tokenize does not roundtrip {{ after \n

[wyattscarpenter]

Bug report

Bug description:

import tokenize, io
source_code = r'''
f"""{80 * '*'}\n{{test}}{{details}}{{test2}}\n{80 * '*'}"""
'''
tokens = tokenize.generate_tokens(io.StringIO(source_code).readline)
x = tokenize.untokenize((t,s) for t, s, *_ in tokens)
print(x)

Expected:

f"""{80 *'*'}\n{{test}}{{details}}{{test2}}\n{80 *'*'}"""

Got:

f"""{80 *'*'}\n{test}}{{details}}{{test2}}\n{80 *'*'}"""

Note the absence of a second { in the {{ after the \n — but in no other positions.

Unlike some other roundtrip failures of tokenize, some of which are minor infelicities, this one actually creates a syntactically invalid program on roundtrip, which is quite bad. You get a SyntaxError: f-string: single '}' is not allowed when trying to use the results.

CPython versions tested on:

3.12

Operating systems tested on:

Linux, Windows

Linked PRs

• gh-125008: Fix tokenize.untokenize roundtrip for \n{{ #125013
• [3.13] gh-125008: Fix tokenize.untokenize roundtrip for \n{{ (GH-125013) #125020
• [3.12] gh-125008: Fix tokenize.untokenize roundtrip for \n{{ (GH-125013) #125021

[wyattscarpenter]

Furthermore, here is the output of the following code:

import tokenize, io
source_code = r'f"\n{{test}}"'
tokens = tokenize.generate_tokens(io.StringIO(source_code).readline)
for t in tokens:
 print(t)

TokenInfo(type=61 (FSTRING_START), string='f"', start=(1, 0), end=(1, 2), line='f"\\n{{test}}"')
TokenInfo(type=62 (FSTRING_MIDDLE), string='\\n{', start=(1, 2), end=(1, 5), line='f"\\n{{test}}"')
TokenInfo(type=62 (FSTRING_MIDDLE), string='test}', start=(1, 6), end=(1, 11), line='f"\\n{{test}}"')
TokenInfo(type=63 (FSTRING_END), string='"', start=(1, 12), end=(1, 13), line='f"\\n{{test}}"')
TokenInfo(type=4 (NEWLINE), string='', start=(1, 13), end=(1, 14), line='f"\\n{{test}}"')
TokenInfo(type=0 (ENDMARKER), string='', start=(2, 0), end=(2, 0), line='')

So, it seems that the line is getting in alright, but the \n{{ is getting turned into a \n{ in the tokenizer somehow.

Same erroneous output for the bytes version (with rb-string, BytesIO and tokenize.tokenize).

[AlexWaygood]

It looks like this was a regression in Python 3.12; I can't reproduce the behaviour with Python 3.11. I'm guessing it was caused by the PEP-701 changes.

[AlexWaygood]

Reproduced on the main branch as well.

[tomasr8]

This seems to happen with other escape characters as well:

import tokenize, io
source_code = r'f"""\t{{test}}"""'
tokens = tokenize.generate_tokens(io.StringIO(source_code).readline)
x = tokenize.untokenize((t,s) for t, s, *_ in tokens)
print(x) # f"""\t{test}}"""

import tokenize, io
source_code = r'f"""\r{{test}}"""'
tokens = tokenize.generate_tokens(io.StringIO(source_code).readline)
x = tokenize.untokenize((t,s) for t, s, *_ in tokens)
print(x) # f"""\r{test}}"""

[tomasr8]

I think the issue is in this method:

cpython/Lib/tokenize.py

Lines 187 to 208 in 16cd6cc

	def escape_brackets(self, token):
	characters = []
	consume_until_next_bracket = False
	for character in token:
	if character == "}":
	if consume_until_next_bracket:
	consume_until_next_bracket = False
	else:
	characters.append(character)
	if character == "{":
	n_backslashes = sum(
	1 for char in _itertools.takewhile(
	"\\".__eq__,
	characters[-2::-1]
	)
	)
	if n_backslashes % 2 == 0:
	characters.append(character)
	else:
	consume_until_next_bracket = True
	characters.append(character)
	return "".join(characters)

This PR fixed the handling of Unicode literals (e.g. \\N{foo}), but it seems to only
be checking for the presence of backslashes without checking if they are followed by N. This appears to fix that:

 if character == "{":
 n_backslashes = sum(
 1 for char in _itertools.takewhile(
 "\\".__eq__,
 characters[-2::-1]
 )
 )
- if n_backslashes % 2 == 0:
+ if n_backslashes % 2 == 0 or characters[-1] != "N":
 characters.append(character)
 else:
 consume_until_next_bracket = True