how to join 8x8bit input into 1x64bit output

[Altera_Forum]

Hi, 

I'm a total VHDL noob, and I feel this question is very easy, but I'm a bit under time pressure and don't want to waste too much time implementing something that is not the correct way how to do it! So far I did some shallow tutorials and, coming from software, I learned that any HDL is rather about state machine implementations, than sequential coding.   I'm playing now with the AVALON MM interface and a Cyclon V SoC FPGA board. In my C code, I have 8 bit chars which I send via /dev/mem (Linux) to the FPGA. On the other side, the FPGA puts them (writedata and write signal) into 8 bit std_logic_vectors. Here another component now needs 8x8bit chunks concatenated to a 1x64bit std_logic_vector. What is a clean VHDL way to get 8x8bit chunks from one input concatenated, and put the 1x64bit std_logic_vector into the other component's input?

[Altera_Forum]

Use SignalTap II and trace the traffic on the Avalon-MM interface you are writing to. Those traces will show you what the transactions look like at the hardware level. 

 Just because you send eight 8-bit chars to /dev/mem does not mean they show up in the Avalon-MM space as 8-bit writes, they may show up as wider writes (with byte-enables set appropriately).  If you want to use standard components, then writing the 8-bit values to RAM, and then programming an Avalon-MM DMA controller to perform a 64-bit read from that same location, and a write to another location may be a suitable solution, i.e., create a Qsys component with RAM, a DMA controller, and your component that requires 64-bit transactions.  Another option would be to; create a component that looks like an 8-bit Avalon-MM slave on one side, and a 64-bit Avalon-MM master on the other. Have the 64-bit side generate a 64-bit transaction each time the 8-bit side has accepted 8 x 8-bit writes (eg., when the write-enable occurs for the last of the 8 bytes). But then you'd have to create HDL code, which might be a little too much at this point ... but if you're willing to learn, and this is the appropriate solution, give it a shot.  Cheers, Dave