[PyPy-issue] [issue515] __future__ import fails (silently) if after a r"string"

[Matteo Bertini]

New submission from Matteo Bertini <matteo at naufraghi.net>:

[11:35] matteo at spovest:~/Downloads/pypy-1.2-osx$ python2.5 with1.py 
"""
something
"""

from __future__ import with_statement

with open(__file__) as stream:
    print "".join(stream)

[11:35] matteo at spovest:~/Downloads/pypy-1.2-osx$ python2.5 with2.py 
r"""
something
"""

from __future__ import with_statement

with open(__file__) as stream:
    print "".join(stream)

[11:35] matteo at spovest:~/Downloads/pypy-1.2-osx$ pypy with1.py 
"""
something
"""

from __future__ import with_statement

with open(__file__) as stream:
    print "".join(stream)

[11:35] matteo at spovest:~/Downloads/pypy-1.2-osx$ pypy with2.py 
Traceback (most recent call last):
  File "?", line 33, in run_toplevel
  File "with2.py", line 7
     with open(__file__) as stream:
         ^
 SyntaxError: invalid syntax

----------
effort: ???
messages: 1678
nosy: hpk, naufraghi, pypy-issue
priority: bug
release: ???
status: unread
title: __future__ import fails (silently) if after a r"string"

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PyPy development tracker <pypy-dev-issue at codespeak.net>
<https://codespeak.net/issue/pypy-dev/issue515>
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