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socket.if_indextoname()

socket.if_indextoname(if_index)

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

Availability: Unix.

New in version 3.3.

Links:

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

doc_python

2016-10-07 17:42:15

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