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(LibAnet) How I Get the ipv6 address from a accept_connection newsocket?
To
:
debian-ada@lists.debian.org
Subject
: (LibAnet) How I Get the ipv6 address from a accept_connection newsocket?
From
: Dani <
danielcheagle@gmail.com
>
Date
: Tue, 5 Sep 2017 15:26:01 -0300
Message-id
: <
[🔎]
CAEXvJLfj548M1AGua84txOmmr3-8rW-eKumR2=qCa3bKPkrh-w@mail.gmail.com
>
Hi!
How I Get the ipv6 address from a accept_connection newsocket?
example case:
Obs:
show_address is just an example.
What I want is what to use instead of this show_address.
for now
I do not know what to use or do.
procedure example is
soq : TCPv6_Socket_Type;
soq_new : TCPv6_Socket_Type;
begin
soq.Init;
soq.Bind (Port => 59000 );
soq.Listen (Backlog => 80 );
soq.Accept_Connection(soq_new); -- bloq
put_line("ipv6 Address from connecting host is " & show_addres(soq_new);
end example;
--
"There are many plans in the Human heart, But
is the Lord's Purpose that prevails"
"Existem Muitos planos e desejos no coração Humano, MAS
são os Propósitos do Senhor que prevalecem"
[]'s Dani:-)
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