mirror of
https://github.com/xvzc/SpoofDPI.git
synced 2024-12-22 22:36:53 +00:00
README.md: Fix typos (#206)
This commit is contained in:
parent
0699e50d15
commit
efa64ab4c0
@ -55,13 +55,13 @@ google-chrome --proxy-server="http://127.0.0.1:8080"
|
|||||||
|
|
||||||
# How it works
|
# How it works
|
||||||
### HTTP
|
### HTTP
|
||||||
Since most of websites in the world now support HTTPS, SpoofDPI doesn't bypass Deep Packet Inspections for HTTP requets, However It still serves proxy connection for all HTTP requests.
|
Since most websites in the world now support HTTPS, SpoofDPI doesn't bypass Deep Packet Inspections for HTTP requests, However, it still serves proxy connection for all HTTP requests.
|
||||||
|
|
||||||
### HTTPS
|
### HTTPS
|
||||||
Although TLS encrypts every handshake process, the domain names are still shown as plaintext in the Client hello packet.
|
Although TLS encrypts every handshake process, the domain names are still shown as plaintext in the Client hello packet.
|
||||||
In other words, when someone else looks on the packet, they can easily guess where the packet is headed to.
|
In other words, when someone else looks on the packet, they can easily guess where the packet is headed to.
|
||||||
The domain name can offer a significant information while DPI is being processed, and we can actually see that the connection is blocked right after sending Client hello packet.
|
The domain name can offer significant information while DPI is being processed, and we can actually see that the connection is blocked right after sending Client hello packet.
|
||||||
I had tried some ways to bypass this, and found out that it seemed like only the first chunk gets inspected when we send the Client hello packet splited in chunks.
|
I had tried some ways to bypass this and found out that it seemed like only the first chunk gets inspected when we send the Client hello packet split into chunks.
|
||||||
What SpoofDPI does to bypass this is to send the first 1 byte of a request to the server,
|
What SpoofDPI does to bypass this is to send the first 1 byte of a request to the server,
|
||||||
and then send the rest.
|
and then send the rest.
|
||||||
|
|
||||||
|
Loading…
Reference in New Issue
Block a user