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socks_proxy's Issues

Failed to receive SOCKS5 connect request ack.

So I've mounted the proxy on port 1080,
Great tool by the way, then on requesting Google,

curl --ipv4 --socks5 127.0.0.1:1080 https://www.google.com --verbose

I've got:

*   Trying 127.0.0.1:1080...
* SOCKS5 connect to IPv6 2800:3f0:4005:407::2004:443 (locally resolved)
* Failed to receive SOCKS5 connect request ack.
* Closing connection 0
curl: (97) Failed to receive SOCKS5 connect request ack.

Does anyone have any clue?

[showcase] a build once run anywhere binary

demo

prebuilt binary: main.com.gz
supported: Linux, NetBSD, FreeBSD, OpenBSD, Mac OS X, Windows

2,19d1
< #include <sys/types.h>
< #include <stdio.h>
< #include <stdarg.h>
< #include <time.h>
< #include <errno.h>
< #include <unistd.h>
< #include <signal.h>
< #include <stdlib.h>
< #include <stdint.h>
< #include <string.h>
< #include <sys/socket.h>
< #include <sys/fcntl.h>
< #include <sys/stat.h>
< #include <netdb.h>
< #include <sys/select.h>
< #include <arpa/inet.h>
< #include <netinet/tcp.h>
< #include <pthread.h>
110c92,93
< 				buf += nread;
---
> 				char* buf_c = (char*)buf;
> 				buf_c += nread;
130c113,114
< 				buf += nwrite;
---
> 				char* buf_c = (char*)buf;
> 				buf_c += nwrite;

Bro

Just want to say sank you!

Segmentation fault

Segmentation fault, if the domain name isn't null-terminated string (SOCKS4a protocol).

#!/bin/python3

import socket

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM, 0)

s.connect(('127.0.0.1', 1080))

s.send(b'\x04\x01\x00\x00\x00\x00\x00\x01\x00')
for i in range(256):
    s.send(b'\xFF')

Add SECURITY.md

Hello ๐Ÿ‘‹

I run a security community that finds and fixes vulnerabilities in OSS. A researcher (@acquykhud) has found a potential issue, which I would be eager to share with you.

Could you add a SECURITY.md file with an e-mail address for me to send further details to? GitHub recommends a security policy to ensure issues are responsibly disclosed, and it would help direct researchers in the future.

Looking forward to hearing from you ๐Ÿ‘

(cc @huntr-helper)

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