Actually you want to call ByteBuf.retainedDuplicate() as you also need to 
increment the reference count.


> On 12. Apr 2019, at 08:19, Frederic <[email protected]> wrote:
> 
> You can create one buffer and then use to send to all your sockets the very 
> same buffer content but using different indexes (read here) using the 
> ByteBuf.duplicate() method.
> 
> https://netty.io/4.1/api/io/netty/buffer/ByteBuf.html#duplicate--
> 
> Javadoc says:
> 
> public abstract ByteBuf 
> <https://netty.io/4.1/api/io/netty/buffer/ByteBuf.html> duplicate()
> Returns a buffer which shares the whole region of this buffer. Modifying the 
> content of the returned buffer or this buffer affects each other's content 
> while they maintain separate indexes and marks. This method does not modify 
> readerIndex or writerIndex of this buffer.
> The reader and writer marks will not be duplicated. Also be aware that this 
> method will NOT call retain() 
> <https://netty.io/4.1/api/io/netty/buffer/ByteBuf.html#retain--> and so the 
> reference count will NOT be increased.
> 
> Returns:
> A buffer whose readable content is equivalent to the buffer returned by 
> slice() <https://netty.io/4.1/api/io/netty/buffer/ByteBuf.html#slice-->. 
> However this buffer will share the capacity of the underlying buffer, and 
> therefore allows access to all of the underlying content if necessary.
> 
> Le vendredi 12 avril 2019 06:31:57 UTC+2, [email protected] a écrit :
> Are there any optimizations available if I'd like to send the very same data 
> to dozens (or hundreds) of Websocket channels?
> At a minimum I'd like to avoid creating (and copying) the data to new 
> buffers. Ideally I would provide a single buffer and a list of channels.
> 
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