[3.12] pythongh-125553: Fix backslash continuation in untokenize (pythonGH-126010) (python#130579)

(cherry picked from commit 7ad793e)

Co-authored-by: Tomas R. <[email protected]>

Author: hugovk

Lib/test/test_tokenize.py

@@ -1,20 +1,20 @@
-from test import support
-from test.support import os_helper
+import os
+import re
+import token
+import unittest
 from tokenize import (tokenize, untokenize, NUMBER, NAME, OP,
                      STRING, ENDMARKER, ENCODING, tok_name, detect_encoding,
                      open as tokenize_open, Untokenizer, generate_tokens,
                      NEWLINE, _generate_tokens_from_c_tokenizer, DEDENT, TokenInfo,
                      TokenError)
 from io import BytesIO, StringIO
-import unittest
 from textwrap import dedent
 from unittest import TestCase, mock
+from test import support
 from test.test_grammar import (VALID_UNDERSCORE_LITERALS,
                                INVALID_UNDERSCORE_LITERALS)
 from test.support import os_helper
 from test.support.script_helper import run_test_script, make_script, run_python_until_end
-import os
-import token
 
 # Converts a source string into a list of textual representation
 # of the tokens such as:
@@ -1816,6 +1816,22 @@ def test_iter_compat(self):
         self.assertEqual(untokenize(iter(tokens)), b'Hello ')
 
 
+def contains_ambiguous_backslash(source):
+    """Return `True` if the source contains a backslash on a
+    line by itself. For example:
+
+    a = (1
+        \\
+    )
+
+    Code like this cannot be untokenized exactly. This is because
+    the tokenizer does not produce any tokens for the line containing
+    the backslash and so there is no way to know its indent.
+    """
+    pattern = re.compile(br'\n\s*\\\r?\n')
+    return pattern.search(source) is not None
+
+
 class TestRoundtrip(TestCase):
 
     def check_roundtrip(self, f):
@@ -1826,6 +1842,9 @@ def check_roundtrip(self, f):
         tokenize.untokenize(), and the latter tokenized again to 2-tuples.
         The test fails if the 3 pair tokenizations do not match.
 
+        If the source code can be untokenized unambiguously, the
+        untokenized code must match the original code exactly.
+
         When untokenize bugs are fixed, untokenize with 5-tuples should
         reproduce code that does not contain a backslash continuation
         following spaces.  A proper test should test this.
@@ -1849,6 +1868,13 @@ def check_roundtrip(self, f):
         tokens2_from5 = [tok[:2] for tok in tokenize(readline5)]
         self.assertEqual(tokens2_from5, tokens2)
 
+        if not contains_ambiguous_backslash(code):
+            # The BOM does not produce a token so there is no way to preserve it.
+            code_without_bom = code.removeprefix(b'\xef\xbb\xbf')
+            readline = iter(code_without_bom.splitlines(keepends=True)).__next__
+            untokenized_code = untokenize(tokenize(readline))
+            self.assertEqual(code_without_bom, untokenized_code)
+
     def check_line_extraction(self, f):
         if isinstance(f, str):
             code = f.encode('utf-8')

Lib/tokenize.py

@@ -171,21 +171,36 @@ def __init__(self):
         self.prev_row = 1
         self.prev_col = 0
         self.prev_type = None
+        self.prev_line = ""
         self.encoding = None
 
     def add_whitespace(self, start):
         row, col = start
         if row < self.prev_row or row == self.prev_row and col < self.prev_col:
             raise ValueError("start ({},{}) precedes previous end ({},{})"
                              .format(row, col, self.prev_row, self.prev_col))
-        row_offset = row - self.prev_row
-        if row_offset:
-            self.tokens.append("\\\n" * row_offset)
-            self.prev_col = 0
+        self.add_backslash_continuation(start)
         col_offset = col - self.prev_col
         if col_offset:
             self.tokens.append(" " * col_offset)
 
+    def add_backslash_continuation(self, start):
+        """Add backslash continuation characters if the row has increased
+        without encountering a newline token.
+
+        This also inserts the correct amount of whitespace before the backslash.
+        """
+        row = start[0]
+        row_offset = row - self.prev_row
+        if row_offset == 0:
+            return
+
+        newline = '\r\n' if self.prev_line.endswith('\r\n') else '\n'
+        line = self.prev_line.rstrip('\\\r\n')
+        ws = ''.join(_itertools.takewhile(str.isspace, reversed(line)))
+        self.tokens.append(ws + f"\\{newline}" * row_offset)
+        self.prev_col = 0
+
     def escape_brackets(self, token):
         characters = []
         consume_until_next_bracket = False
@@ -245,8 +260,6 @@ def untokenize(self, iterable):
                     end_line, end_col = end
                     extra_chars = last_line.count("{{") + last_line.count("}}")
                     end = (end_line, end_col + extra_chars)
-            elif tok_type in (STRING, FSTRING_START) and self.prev_type in (STRING, FSTRING_END):
-                self.tokens.append(" ")
 
             self.add_whitespace(start)
             self.tokens.append(token)
@@ -255,6 +268,7 @@ def untokenize(self, iterable):
                 self.prev_row += 1
                 self.prev_col = 0
             self.prev_type = tok_type
+            self.prev_line = line
         return "".join(self.tokens)
 
     def compat(self, token, iterable):

Misc/NEWS.d/next/Library/2024-10-26-16-59-02.gh-issue-125553.4pDLzt.rst

@@ -0,0 +1,2 @@
+Fix round-trip invariance for backslash continuations in
+:func:`tokenize.untokenize`.