BUG: read_csv not raising when \n as sep (#44899)

Author: phofl

doc/source/whatsnew/v1.4.0.rst

@@ -749,6 +749,7 @@ I/O
 - Bug in :func:`read_csv` raising ``ValueError`` when names was longer than header but equal to data rows for ``engine="python"`` (:issue:`38453`)
 - Bug in :class:`ExcelWriter`, where ``engine_kwargs`` were not passed through to all engines (:issue:`43442`)
 - Bug in :func:`read_csv` raising ``ValueError`` when ``parse_dates`` was used with ``MultiIndex`` columns (:issue:`8991`)
+- Bug in :func:`read_csv` not raising an ``ValueError`` when ``\n`` was specified as ``delimiter`` or ``sep`` which conflicts with ``lineterminator`` (:issue:`43528`)
 - Bug in :func:`read_csv` converting columns to numeric after date parsing failed (:issue:`11019`)
 - Bug in :func:`read_csv` not replacing ``NaN`` values with ``np.nan`` before attempting date conversion (:issue:`26203`)
 - Bug in :func:`read_csv` raising ``AttributeError`` when attempting to read a .csv file and infer index column dtype from an nullable integer type (:issue:`44079`)

pandas/io/parsers/readers.py

@@ -1460,6 +1460,13 @@ def _refine_defaults_read(
             "delim_whitespace=True; you can only specify one."
         )
 
+    if delimiter == "\n":
+        raise ValueError(
+            r"Specified \n as separator or delimiter. This forces the python engine "
+            "which does not accept a line terminator. Hence it is not allowed to use "
+            "the line terminator as separator.",
+        )
+
     if delimiter is lib.no_default:
         # assign default separator value
         kwds["delimiter"] = delim_default

pandas/tests/io/parser/common/test_common_basic.py

@@ -788,6 +788,22 @@ def test_read_csv_delimiter_and_sep_no_default(all_parsers):
         parser.read_csv(f, sep=" ", delimiter=".")
 
 
+@pytest.mark.parametrize("kwargs", [{"delimiter": "\n"}, {"sep": "\n"}])
+def test_read_csv_line_break_as_separator(kwargs, all_parsers):
+    # GH#43528
+    parser = all_parsers
+    data = """a,b,c
+1,2,3
+    """
+    msg = (
+        r"Specified \\n as separator or delimiter. This forces the python engine "
+        r"which does not accept a line terminator. Hence it is not allowed to use "
+        r"the line terminator as separator."
+    )
+    with pytest.raises(ValueError, match=msg):
+        parser.read_csv(StringIO(data), **kwargs)
+
+
 def test_read_csv_posargs_deprecation(all_parsers):
     # GH 41485
     f = StringIO("a,b\n1,2")