functools.partial as argument to _out fails

[ianw]

Something like

import functools
import sh

def _log_line(foo, line, stdin, process):
    print "%s : %s" % (foo, line)

log_line = functools.partial(_log_line, "hello")

a = sh.ls("-l", "/var/tmp", _out=log_line)

fails with

Traceback (most recent call last):
  File "/tmp/test.py", line 9, in <module>
    a = sh.ls("-l", "/var/tmp", _out=log_line)
  File "/opt/data/env/lib/python2.7/site-packages/sh.py", line 769, in __call__
    return RunningCommand(cmd, call_args, stdin, stdout, stderr)
  File "/opt/data/env/lib/python2.7/site-packages/sh.py", line 327, in __init__
    self.call_args, pipe=pipe)
  File "/opt/data/env/lib/python2.7/site-packages/sh.py", line 961, in __init__
    save_data=save_stdout)
  File "/opt/data/env/lib/python2.7/site-packages/sh.py", line 1360, in __init__
    num_args = len(inspect.getargspec(handler.__call__).args)
  File "/usr/lib64/python2.7/inspect.py", line 815, in getargspec
    raise TypeError('{!r} is not a Python function'.format(func))
TypeError: <method-wrapper '__call__' of functools.partial object at 0x7f31bada11b0> is not a Python function

It seems to be a semi-common problem with inspect and functools. There are several pointers to the alternative implementation

https://bitbucket.org/macfreek/sphinx/src/2db7eac2830b/sphinx/util/inspect.py#cl-24

when googling for the error.

[amoffat]

Thanks for reporting this Ian. It's a result of sh trying to be too clever...I will take a look at this soon.

[amoffat]

fixed on release-1.10 branch