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socket.if_nametoindex()

socket.if_nametoindex(if_name)

Return a network interface index number corresponding to an interface name. OSError if no interface with the given name exists.

Availability: Unix.

New in version 3.3.

Links:

https://docs.python.org/3.5/library/socket.html#socket.if_nametoindex

doc_python

2016-10-07 17:42:16

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