so recently I created a Tunnel Broker and follow their configuration, and follow this configuration to assign IPv6 addresses to my host Configuration
When I disable the IPv4 address to my pc it won’t connect to the internet but still able to ping to Tunnel servers and google dns so I try to enable it back and it works fine. When I try to ping from my pc to a website it will show their IPv6.
Here is my address list and allocated ip IP Address
so I want to ask can I connect to the internet by using an IPv6 address only or I still need an IPv4 allocated to my host/pc to be able to connect?
Using an IPv6 address you can only connect to the IPv6-enabled part of the internet.
So you will be able to use websites like Google or Facebook or Youtube without problems, but there will be a large percentage of websites that you cannot visit because they have only IPv4.
So the answer is: it depends on what you want to do.
I would say in a realistic view, a dual-stack (both IPv4 and IPv6) is the only realistic solution for now, and probably at least the next 10 years.
Also, a tunnel broker is just a detour in all your traffic, like a VPN is. You need to get IPv6 from your own ISP for it to be really effective.
Yeah I just notice it since My IPv6 will only work when the router still enables the IPv4 address.
Are there any references that I can read about this matter? books or papers?
This is normal when you use a tunnel because the IPv6 traffic is sent within IPv4 traffic to your tunnel broker.
When you have an ISP that provides you with a direct IPv6 address and you configure that on your router, you will not have this problem.
But as I wrote, using only IPv6 is not yet practical.
If you’d want to go further into future, you can have your network IPv6-only with the help of DNS64/NAT64. But RouterOS can’t help you with that, you’d need another device. There’s really no practical advantage it would give you (the opposite would be true), but it could be interesting as a fun thing to play with (on some indepedent experimental network segment).