[issue21295] Python 3.4 gives wrong col_offset for Call nodes returned from ast.parse
Radek Novacek added the comment:
There is still problem with col_offset is some situations, for example
col_offset of the ast.Attribute should be 4 but is 0 instead:
>>> for x in ast.walk(ast.parse('foo.bar')):
... if hasattr(x, 'col_offset'):
... print("%s: %d" % (x, x.col_offset))
...
<_ast.Expr object at 0x7fcdc84722b0>: 0
<_ast.Attribute object at 0x7fcdc84723c8>: 0
<_ast.Name object at 0x7fcdc8472438>: 0
Is there any solution to this problem? It causes problems in python support in
KDevelop (kdev-python).
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[issue21295] Python 3.4 gives wrong col_offset for Call nodes returned from ast.parse
Radek Novacek added the comment:
Aivar, I have to admit that my knowledge of this is limited, but as I
understand it, the attribute is "bar" in the "foo.bar" expression.
I can get beginning of the assignment by
>>> ast.parse('foo.bar').body[0].value.value.col_offset
0
But how can I get position of the 'bar'? My guess is this:
>>> ast.parse('foo.bar').body[0].value.col_offset
but it still returns 0.
Why this two col_offsets returns the same value? How can I get the position of
'bar' in 'foo.bar'?
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[issue21295] Python 3.4 gives wrong col_offset for Call nodes returned from ast.parse
Radek Novacek added the comment:
I've ran the tests from first and second comment using python 3.5.0 and it
seems it produces correct results:
>>> import ast
>>> tree = ast.parse("sin(0.5)")
>>> first_stmt = tree.body[0]
>>> call = first_stmt.value
>>> print("col_offset of call expression:", call.col_offset)
col_offset of call expression: 0
>>> print("col_offset of func of the call:", call.func.col_offset)
col_offset of func of the call: 0
>>> tree = ast.parse("(sin\n(0.5))")
>>> first_stmt = tree.body[0]
>>> call = first_stmt.value
>>> print("col_offset of call expression:", call.col_offset)
col_offset of call expression: 1
>>> print("col_offset of func of the call:", call.func.col_offset)
col_offset of func of the call: 1
>>> print("lineno of call expression:", call.lineno)
lineno of call expression: 1
>>> print("lineno of func of the call:", call.lineno)
lineno of func of the call: 1
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nosy: +rnovacek
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