how_to_use.txt

这是一个LINUX环境的 VDR 插件源代码,可支持Irdeto, Seca, Viaccess, Nagra, Conax & Cryptoworks等CA系统的读卡、共享等操作。

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FFdecsa
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This code is able to decrypt MPEG TS packets with the CSA algorithm. To
achieve high speed, the decryption core works on many packets at the
same time, so the interface is more complicated than usual decsa
implementations.

The FFdecsa.h file defines the external interface of this code.

Basically:

1) you use get_suggested_cluster_size to know the optimal number of
packets you have to pass for decryption

2) you use set_control_words to set the decryption keys

3) you use decrypt_packets to do the actual decryption

You don't need to always use set_control_words before decrypt_packets,
if keys aren't changed. 


The decrypt_packets function call decrypts many packets at the same
time. The interface is complicated because the only design goal was
speed, so it implements zero-copying of packets, out-of-order decryption
and optimal packet aggregation for better parallelism. This part is the
most difficult to understand.

--- HOW TO USE int decrypt_packets(unsigned char **cluster); ---

PARAMETERS
  cluster points to an array of pointers, representing zero or more
  ranges. Every range has a start and end pointer; a start pointer==NULL
  terminates the array.
  So, an array of pointers has this content:
    start_of_buffer_1, end_of_buffer_1, ... start_of_buffer_N,
    end_of_buffer_N, NULL
  example:
    0x12340000, 0x123400bc, 0x56780a00, 0x5678b78, NULL
  has two ranges (0x12340000 - 0x123400bc and  0x56780a00 - 0x5678b78),
  for a total of three packets (starting at 0x12340000, 0x56780a00,
  0x5678abc)
RETURNS
  How many packets can now be consumed by the caller, this is always >=
  1, unless the cluster contained zero packets (in that case it's
  obviously zero).
MODIFIES
  The cluster is modified to try to exclude packets which shouldn't be
  submitted again for decryption (because just decrypted or originally
  not crypted). "Try to exclude" because the returned array will never
  be bigger than what was passed, so if you passed only a range and some
  packets in the middle were decrypted making "holes" into the range,
  the range would have to be split into several ranges, and that will
  not be done. If you want a strict description of what has to be passed
  again to decrypt_packets, you have to use ranges with only one packet
  inside. Note that the first packet will certainly be eliminated from
  the returned cluster (see also RETURNS).

You can now read the detailed description of operation or just skip to
the API examples.


---------------------------------
DETAILED DESCRIPTION OF OPERATION
---------------------------------
  consider a sequence of packets like this:
   0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
   E  E  E  E  E  E  E  E  E  E  E  O  E  O  E  O  O  0  0  0  0  0  0  0  0  c  O  O  O  O  O  O  O  O  O  O  O ...
  where
   E = encrypted_even,
   O = encrypted_odd,
   e = clear_was_encrypted_even,
   o = clear_was_encrypted_odd,
   c = clear
  and suppose the suggested cluster size is 10 (this could be for a function with internal parallelism 8)

  1) we define the cluster to include packets 0-9 and
  call decrypt_packets
  a possible result is that the function call
  - returns 8 (8 packets available)
  - the buffer contains now this
  -----------------------------
   0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
   e  e  e  e  e  e  e  e  E  E  E  O  E  O  E  O  O  0  0  0  0  0  0  0  0  c  O  O  O  O  O  O  O  O  O  O  O ...
                          -----
  - the modified cluster covers 8-9 [continue reading, but then see note 1 below]
  so, we can use the first 8 packets of the original cluster (0-7)

  2) now, we define cluster over 8-17 and call decrypt_packets
  a possible result is:
  - returns 3 (3 packets available)
  - the buffer contains now this (!!!)
                          -----------------------------
   0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
   e  e  e  e  e  e  e  e  e  e  e  O  e  O  e  O  O  0  0  0  0  0  0  0  0  c  O  O  O  O  O  O  O  O  O  O  O ...
                                   --    --    --------
  - the modified cluster covers 11-11,13-13,15-17 [continue reading, but then see note 1 below]
  so, we can use the first 3 packets of the original cluster (8-10)

  3) now, we define cluster over 11-20 and call decrypt packets (defining a cluster 11-11,13-13,15-22 would be better)
  a possible result is:
  - returns 10 (10 packets available)
  - the buffer contains now this
                                   -----------------------------
   0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
   e  e  e  e  e  e  e  e  e  e  e  o  e  o  e  o  o  o  o  o  o  0  0  0  0  c  O  O  O  O  O  O  O  O  O  O  O ...

  - the modified cluster is empty
  so, we can use the first 10 packets of the original cluster (11-20)
  What it happened is that the second call decrypted packets 12 and 14 but they were
  not made available because packet 11 was still encrypted,
  the third call decrypted 11,13,15-20 and included 12 and 14 as available too.

  4) now, we define cluster over 21-30 and call decrypt packets
  a possible result is:
  - returns 9 (9 packets available)
  - the buffer contains now this
                                                                 -----------------------------
   0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
   e  e  e  e  e  e  e  e  e  e  e  o  e  o  e  o  o  o  o  o  o  o  o  o  o  c  o  o  o  o  O  O  O  O  O  O  O ...
                                                                                            --
  - the modified cluster covers 30-30
  so, we can use the first 9 packets of the original cluster (21-29)
  What happened is that packet 25 could be skipped because it is in clear.

  Note that the suggested cluster size (10) is higher than the maximum number
  of packets that can be really decrypted (8), but we are able to skip 12 and 14
  in step 3) and run the decryption on a full 8 packets group.
  In the same way, we were able to skip 25 in step 4).
  There are three kinds of "free" packets we can skip:
  - packets decrypted in a previous call (as 12 and 14)
  - packets already in clear (as 25)
  - packets with a payload of less than 8 bytes (clear==encrypted!)

  Note also that we could have defined a better cluster in step 3
  (11-11,13-13,15-22), using what step 2 had returned. The risk of not
  having 8 packets to decrypt would have been smaller (consider the case
  where 19 and 20 were "c").

  Final considerations:
  - you can use a bigger or smaller cluster than the suggested number of packets
  - every call to decrypt_packets has a *fixed* CPU cost, so you should try to
    not run it with a few packets, when possible
  - decrypt_packets can't decrypt even and odd at the same time; it guarantees
    that the first packet will be decrypted and tries to decrypt as many packets
    as possible
  - clear packets in the middle of encrypted packets don't happen in real world,
    but E,E,E,O,E,O,O,O sequences do happen (audio/video muxing problems?) and
    small packets (<8 bytes) happen frequently; the ability to skip is useful.

  note 1:
    As the returned cluster will not have more ranges than the passed one, what it is
    described above is not actually true.
    In the step 1) the returned cluster will cover 8-9, but in step 2) it will
    cover 11-17 (some extra packets had to remain in); this lack of information
    prevents us from using an optimal 11-11,13-13,15-22 in step 3). Note that
    in any case step 3) will decrypt 11,13,15,16,17,18,19,20 thanks to the
    extra margin we use (we put ten packets (including 19 and 20) even if the
    parallelism was just 8, and it was a good idea; but if 19 and 20 were of
    type c, we would have run the decryption with only 6/8 efficiency).
    This problem can be prevented by using ranges with only one packet: in
    step 2) we would have passed
    8-8,9-9,10-10,11-11,12-12,13-13,14-14,15-15,16-16,17-17
    and got back
    11-11,13-13,15-17.


------------
API EXAMPLES
------------

Some examples of how the API can be used (this is not real code, so it
may have typos or other bugs).


Example 1: (big linear buffer, simple use of cluster)

  unsigned char *p;
  unsigned char *cluster[3];
  for(p=start;p<end;){
    cluster[0]=p;cluster[1]=end;
    cluster[2]=NULL;
    p+=188*decrypt_packets(cluster);
  }
  //consume(start,end);


Example 2: (circular buffer, simple use of cluster)

  unsigned char *p;
  unsigned char *cluster[5];

  while(1){
    if(read==write){
      //buffer is empty
      //write=refill_buffer(write,start,end);
      continue;
    }
    else if(read<write){
      cluster[0]=read;cluster[1]=write;
      cluster[2]=NULL;
    }
    else{
      cluster[0]=read;cluster[1]=end;
      cluster[2]=start;cluster[3]=write;
      cluster[4]=NULL;
    }
    new_read=read+188*decrypt_packets(cluster);
    if(new_read<=end){
      //consume(read,new_read);
    }
    else{
      new_read=start+(new_read-end);
      //consume(read,end);
      //consume(start,new_read);
    }
    read=new_read;
    if(read==end) read=start;
  }


Example 3: (undefined buffer structure, advanced use of cluster)

  unsigned char *packets[1000000];
  unsigned char *cluster[142]; //if suggested packets is 70
  
  cluster[0]=NULL;
  for(n=0;n<1000000;){
    i=0;
    while(cluster[2*i]!=NULL) i++; //preserve returned ranges
    for(k=i;k<70&&n<1000000;k++,n++){
      cluster[2*k]=packets[n];cluster[2*k+1]=packets[n]+188;
    }
    cluster[2*k]=NULL;
    decrypt_packets(cluster);
  }
  //consume_all_packets();