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Re: What interface is a ‘Popen.stdout’ presenting?

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On Wed, Dec 23, 2015 at 7:36 PM, Ben Finney <ben+python@benfinney.id.au> wrote:
> So how do I get from a Python 2 ‘file’ object, to whatever
> ‘io.TextIOWrapper’ wants?

I would dup the file descriptor and close the original file. Then open
the file descriptor using io.open:

    >>> p = subprocess.Popen(['uname'], stdout=subprocess.PIPE)
    >>> fd = os.dup(p.stdout.fileno())
    >>> fd
    4
    >>> p.stdout.close()
    >>> fout = io.open(fd, 'r')
    >>> fout
    <_io.TextIOWrapper name=4 encoding='UTF-8'>
    >>> fout.read()
    u'Linux\n'

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Re: What interface is a ‘Popen.stdout’ presenting? eryk sun <eryksun@gmail.com> - 2015-12-23 22:30 -0600

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