Something like protoc --decode_raw?
As for manipulating these in code, there's nothing too great available
if you don't have a descriptor at all. You could create a dummy
message (no fields) and then get them via unknown fields in the
message's reflection object, but they're not easy to manipulate. Note
that the encoded information is not always sufficient for correct
decoding. For example, strings, submessages, and perhaps packed arrays
are encoded in the same way. I also don't think that whether zigzag
encoding is used on varints is part of the serialized data. You really
really really want a descriptor. If not compiled in, then passed along
with the data.
-ilia
On Mon, Aug 19, 2013 at 3:37 PM, Tom Ward <tw...@thefoundry.co.uk> wrote:
> Hi all,
>
> I'm just getting to grips with protobuf and, having read through the
> documentation, I'm struggling to find what I'm after.
>
> Basically I'm trying to work out how to deserialize a protobuf message
> without using the generated headers, as we're likely to get messages that
> weren't generated at compile time. I've looked through the documentation,
> but I only seem to be able to find ones that use generated classes to
> deserialize, or that use a Descriptor from a generated class to create a
> DynamicMessage, which I can't seem to work out how to do when we don't have
> the proto.
>
> Here's a very simple example of what I mean, where Message* is set to some
> base type that allows deserialization, and then can use the reflection API
> (or some factory) to correctly deal with the message:
>
> #include <sstream>
>
> #include "google/protobuf/message.h"
>
> int main(int argc, char **argv)
> {
> if( argc > 1)
> {
> // This will fail to compile, as Message is pure abstract
> google::protobuf::Message* message = NULL; //new
> google::protobuf::Message();
> std::istringstream ss(std::string(argv[1],strlen(argv[1])));
>
> message->ParseFromIstream(&ss);
>
> // do something with the message
> // ...
> }
>
> return 0;
> }
>
>
> Apologies in advance if this has already been answered...
>
> Thanks
>
> Tom
>
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