排查rtmp协议推流时握手bug

概况

转推流程序的过程:从一个观看地址拉流,然后推流到另一个推流地址。主要用于cdn之间转推,目前市面上大多数cdn厂商都愿意不支持动态转推,因此只能通过转推流程序进行转推。

bug现象:使用obs studio推流到微赞可以成功,但是使用Erlang版本的转推流程序推流到微赞却失败。

日志如下:

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14:12:35.926 [debug] payload [{amf,["onBWDone",0,null]}], msgtype[command_msg_4_amf0] 
14:12:35.949 [debug] play succ ======> url ["/live-sz/w1520993434573948"]
14:12:35.949 [debug] {rtmp_msg,4,0,data_msg_4_amf0,1,{amf,["|RtmpSampleAccess",true,true]}}
14:12:35.949 [debug] {rtmp_msg,4,0,data_msg_4_amf0,1,{amf,["onStatus",{object,[{"code","NetStream.Data.Start"}]}]}}
14:12:36.038 [error] gen_server <0.122.0> terminated with reason: no match of right hand value <<0,0,0,0,0,0,0,0,113,142,194,240,185,25,41,180,242,33,5,112,128,97,178,8,79,179,28,53,152,242,82,43,234,104,113,246,170,189,182,146,122,36,155,3,152,180,226,122,36,97,52,67,53,158,107,170,178,119,209,132,40,233,102,182,142,233,218,71,55,8,121,67,117,58,130,91,107,224,202,5,1,132,37,245,143,231,20,198,121,204,57,80,102,165,104,245,79,71,254,169,15,3,166,12,148,45,24,62,253,66,93,139,84,139,54,236,47,5,98,95,51,231,222,144,8,153,232,166,227,151,57,98,214,63,238,167,212,49,51,160,83,248,246,199,...>> in rtmp_handshake:create_c2/2 line 61

很显然是rtmp_handshake:create_c2/2函数出现匹配错误,对应代码如下:

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-spec create_c2(C0C1, S0S1S2) -> Result when
C0C1 :: iodata(),
S0S1S2 :: binary(),
Result :: {ok, C2},
C2 :: iolist().
create_c2(C0C1, S0S1S2) when is_list(C0C1) ->
create_c2(iolist_to_binary(C0C1), S0S1S2);
create_c2(<<_C0:1/binary, C1:16#600/binary>>, <<S0:1/binary, S1:16#600/binary, S2:16#600/binary>>) ->
<<3>> = S0,
case C1 of
<<_:32, 0:32, _/binary>> ->
S2 = C1,
{ok, S1};
_ ->
{ok, S1DigestData} = verify_s1(S1),
DigestKey = crypto:hmac(sha256, ?C2_PUBLIC_KEY, S1DigestData),
C2Len = 16#600,
C2DigestDataLen = 32,
RandomBin = random_binary(C2Len - 8 - C2DigestDataLen),
{T1, T2, _} = now(),
Epoch = <<(1000000 * T1 + T2):32/little>>,
Data = [Epoch, binary:part(S1, 0, 4), RandomBin],
S2DigestData = crypto:hmac(sha256, DigestKey, Data),
{ok, [Data, S2DigestData]}
end.

rtmp握手过程中C1数据包匹配<<_:32, 0:32, _/binary>>格式后和S2数据包匹配不成功,程序直接crash dump。因此需要弄清楚rtmp握手过程中是否有对S2和C1进行匹配验证。

Rtmp握手过程

此处重点关注rtmp握手过程中的C1和S2数据包。

先看官方文档中的握手过程,中文翻译版本可以参见:rtmp规范1.0。 官方文档中对于是否要保证C1和S2完全一致,并没有明确说法。因此可以先参见obs studio依赖的librtmp库,看握手过程是如何处理的。obs studio依赖的librtmp的代码如下连接:

  1. https://github.com/obsproject/obs-studio/blob/master/plugins/obs-outputs/librtmp/rtmp.c
  2. https://github.com/obsproject/obs-studio/blob/master/plugins/obs-outputs/librtmp/handshake.h

第一个链接rtmp.c中的代码是推流地址中没有加密串的情况下的握手过程代码,第二个链接handshake.h中的代码是推流地址中有加密串的情况下的握手过程代码。代码中使用条件编译CRYPTO宏来选择编译不同的代码。其中HandShake函数属于客户端的握手函数,SHandShake属于服务端的握手函数。非加密版本具体C语言代码如下(已添加对应的中文注释进行说明):

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#ifndef CRYPTO
static int
HandShake(RTMP *r, int FP9HandShake)
{
//C0,C1 -- S0, S1, S2 -- C2 消息握手协议
int i;
uint32_t uptime, suptime;
int bMatch;
char type;
char clientbuf[RTMP_SIG_SIZE + 1], *clientsig = clientbuf + 1;
char serversig[RTMP_SIG_SIZE];

clientbuf[0] = 0x03; //C0, 一个字节。03代表协议版本号为3 /* not encrypted */

uptime = htonl(RTMP_GetTime()); //这是一个时间戳,放在C1消息头部
memcpy(clientsig, &uptime, 4);

memset(&clientsig[4], 0, 4); //后面放4个字节的空数据然后就是随机数据

//后面是随机数据,总共1536字节的C1消息
#ifdef _DEBUG
for (i = 8; i < RTMP_SIG_SIZE; i++)
clientsig[i] = 0xff;
#else
for (i = 8; i < RTMP_SIG_SIZE; i++)
clientsig[i] = (char)(rand() % 256);
#endif

//发送C0, C1消息
if (!WriteN(r, clientbuf, RTMP_SIG_SIZE + 1))
return FALSE;

//下面读一个字节也就是S0消息,看协议是否一样
if (ReadN(r, &type, 1) != 1) /* 0x03 or 0x06 */
return FALSE;

RTMP_Log(RTMP_LOGDEBUG, "%s: Type Answer : %02X", __FUNCTION__, type);

if (type != clientbuf[0]) //C/S版本不一致
RTMP_Log(RTMP_LOGWARNING, "%s: Type mismatch: client sent %d, server answered %d",
__FUNCTION__, clientbuf[0], type);

//读取S1消息,里面有服务器运行时间
if (ReadN(r, serversig, RTMP_SIG_SIZE) != RTMP_SIG_SIZE)
return FALSE;

/* decode server response */

memcpy(&suptime, serversig, 4);
suptime = ntohl(suptime);

RTMP_Log(RTMP_LOGDEBUG, "%s: Server Uptime : %d", __FUNCTION__, suptime);
RTMP_Log(RTMP_LOGDEBUG, "%s: FMS Version : %d.%d.%d.%d", __FUNCTION__,
serversig[4], serversig[5], serversig[6], serversig[7]);

/* 2nd part of handshake */
if (!WriteN(r, serversig, RTMP_SIG_SIZE)) //发送C2消息,内容就等于S1消息的内容。
return FALSE;

//读取S2消息
if (ReadN(r, serversig, RTMP_SIG_SIZE) != RTMP_SIG_SIZE)
return FALSE;

//服务端返回的S2消息和C1消息进行了比对,但是即使没有match,也是返回TRUE,只是打印了log
bMatch = (memcmp(serversig, clientsig, RTMP_SIG_SIZE) == 0);
if (!bMatch)
{
RTMP_Log(RTMP_LOGWARNING, "%s, client signature does not match!", __FUNCTION__);
}

/* er, totally unused? */
(void)FP9HandShake;
return TRUE;
}

static int
SHandShake(RTMP *r)
{
...
return TRUE;
}
#endif

rtmp握手过程中确实存在对S2和C1进行匹配验证的操作,但是这个操作并不影响握手是否是成功的,只是添加了一条warnning日志而已。因此obs studio还是能推流成功。相对应的在我们的转推流程序中,需要针对这个情况不进行强认证,删除掉匹配的操作即可。

抓包分析

以微赞和网宿为例

网宿推流没有走加密流程,S2和C1匹配,具体数据包截图如下:

obs-netcenter-handshake-success.jpg

微赞推流走加密流程,S2和C1不匹配,具体数据包截图如下:

obs-vzan-handshake-success.jpg

到此,整个rtmp推流握手过程就比较清楚了。

因此只需要将Erlang代码的流程更改下即可(删除S2和C1的匹配过程),见下面的Erlang代码:

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-spec create_c2(C0C1, S0S1S2) -> Result when
C0C1 :: iodata(),
S0S1S2 :: binary(),
Result :: {ok, C2},
C2 :: iolist().
create_c2(C0C1, S0S1S2) when is_list(C0C1) ->
create_c2(iolist_to_binary(C0C1), S0S1S2);
create_c2(<<_C0:1/binary, C1:16#600/binary>>, <<S0:1/binary, S1:16#600/binary, _S2:16#600/binary>>) ->
<<3>> = S0,
case C1 of
<<_:32, 0:32, _/binary>> ->
{ok, S1};
_ ->
{ok, S1DigestData} = verify_s1(S1),
DigestKey = crypto:hmac(sha256, ?C2_PUBLIC_KEY, S1DigestData),
C2Len = 16#600,
C2DigestDataLen = 32,
RandomBin = random_binary(C2Len - 8 - C2DigestDataLen),
{T1, T2, _} = now(),
Epoch = <<(1000000 * T1 + T2):32/little>>,
Data = [Epoch, binary:part(S1, 0, 4), RandomBin],
S2DigestData = crypto:hmac(sha256, DigestKey, Data),
{ok, [Data, S2DigestData]}
end.

至此,转推流成功,示例图如下:

rtmp_deliver_success.jpg

结论

虽然Adobe公司自己出的rtmp协议不是iso标准的,但是你们这些公司好歹也尽量按照规定来啊,贼鸡儿坑。


参考:

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