[ByteBuffer] read<T>() function

[Toinan67]

Hi !

It's been a very long time since i posted, i'm pleased to be back

I have a question about the ByteBuffer class, which is the parent class of WorldPacket.

If i understood correctly, when you do this :

uint32 value;

packet >> value;

You get into that function :

       ByteBuffer &operator>>(uint32 &value)
       {
           value = read<uint32>();
           return *this;
       }

Which ultimately calls that function :

template <typename T> T read(size_t pos) const
       {
           if(pos + sizeof(T) > size())
               throw ByteBufferException(false, pos, sizeof(T), size());
           T val = *((T const*)&_storage[pos]);
           EndianConvert(val);
           return val;
       }

_storage is a vector<uint8>.

I have a problem with :

T val = *((T const*)&_storage[pos]);

(sorry about the code display, it seems like there is some troubles)

So basically we're storing a uint32 in a uint8 ?!

Then why do we use uint32? Why not using uint8 everywhere?

And why is this line so complicated? Why not :

T val = _storage[pos];

I hope i've been clear

Thank you!

[lillecarl]

one byte is 8 bit, i think that might be the reason ^^,

[amaru]

_storage is byte "array" by default. such arrays in memory are located as continuous set of bytes. _storage as value is a pointer to it's first element

if uint8 storage[n]; then storage == &storage[0] which means that storage val points to the head of that array

when accessing _storage[pos] you access *(_storage + pos), which means byte number 'pos' since address of _storage.

by casting *((T const*)&_storage[pos] you say that &_storage[pos] is pointer not to uint8 but to value type T*, and it would 'grab' nearest bytes depending on T size

[Toinan67]

Very interesting, thank you amaru

I had no idea of : "and it would 'grab' nearest bytes depending on T size"

Thanks again