Monday, October 6, 2014

More SystemVerilog Streaming Examples

In my previous post I promised I would write about more interesting cases of streaming using a slice_size and arrays of ints and bytes.  Well, I just posted another set of streaming examples to edaplayground.  I'm going to mostly let you look at that code and learn by example, but I will take some time in this post to explain what I think is the trickiest of the conversions.  When you go to edaplayground, choose the Modelsim simulator to run these.  Riviera-PRO is currently the only other SystemVerilog simulator choice on edaplayground, and it messes up on the tricky ones (more on that in a bit).

These examples demonstrate using the streaming operator to do these conversions:
  • unpacked array of bytes to int
  • queue of ints to queue of bytes
  • queue of bytes to queue of ints
  • int to queue of bytes
  • class to queue of bytes
Of all those examples, the queue of ints to queue of bytes and the queue of bytes to queue of ints are the tricky ones that I want to spend more time explaining.  They are both tricky for the same reason.  If you are like me, your first thought on how to convert a queue of ints to a queue of bytes is to just do this:

byte_queue_dest = {<< byte{int_queue_source}};

Before I explain why that might not be what you want, be sure you remember what "right" and "left" mean from my previous post.  The problem with the straightforward streaming concatenation above is it will start on the right of the int queue (int_queue_source[max_index], because it's unpacked), grab the right-most byte of that int (int_queue_source[max_index][7:0], because the int itself is packed), and put that byte on the left of byte_queue_dest (byte_queue_dest[0], because it is unpacked).  It will then grab the next byte from the right of the int_queue (int_queue_source[max_index][15:8]) and put it in the next position of byte_queue_dest (byte_queue_dest[1]), and so on.  The result is that you end up with the ints from the int queue reversed in the byte queue.  If that doesn't make sense, change the code in the example to the above streaming concatenation and just try it.

To preserve the byte ordering, you do this double streaming concatenation:

byte_queue_dest = {<< byte{ {<< int{int_queue_source}} }};

Let's step through this using the literal representation of the arrays so that rights and lefts will be obvious.  You start with an (unpacked, of course) queue of ints:

int_queue_source = {'h44332211, 'h88776655};

And just to be clear, that means int_queu_source[0][7:0] is 'h11 and we want that byte to end up as byte_queue_dest[0].  The inner stream takes 32-bits at a time from the right and puts them on the left of a temporary packed array of bits.  That ends up looking like this:

temp_bits = 'h8877665544332211;

Now the outer stream takes 8 bits at a time from the right of that and puts them on the left of a queue.  That gives you this in the end:

byte_queue_dest = {'h11, 'h22, 'h33, 'h44, 'h55, 'h66, 'h77, 'h88};

Which, if you wanted to preserve the logical byte ordering, is correct.  Going from bytes to ints, it turns out, is pretty much the same: reverse the queue of bytes and then stream an int at a time.

So what happens with Riviera-PRO?  If you try it in edaplayground you see that the resulting queue of bytes in the int-to-byte conversion ends up with a whole bunch of extra random bytes on the right (highest indexes of the queue).  8 extra, to be exact.  Same for the int queue result in the byte-to-int conversion.  I think Riviera-PROP must be streaming past the end of the temporary packed array (that I called temp_bits above) of pulling in bytes from off in the weeds.  Pretty crazy.  That's all done behind the scenes so I don't really know, but that's sure what it looks like.  Hopefully they can fix that soon.

Well, I hope I've helped clear up how to use the streaming operators for someone.  If I haven't, I have at least helped myself understand them better.  Ask any questions you have in the comments.

Friday, October 3, 2014

SystemVerilog Streaming Operator: Knowing Right from Left

SystemVerilog has this cool feature that is very handy for converting one type of collection of bits into another type of collection of bits.  It's the streaming operator.  Or the streaming concatenation operator.  Or maybe it's the concatenation of streaming expressions (it's also called pack/unpack parenthetically).  Whatever you want to call it, it's nice.  If you have an array of bytes and you want to turn it into an int, or an array of ints that you want to turn into an array of bytes, or if you have a class instance that you want to turn into a stream of bits, then streaming is amazing.  What used to require a mess of nested for-loops can now be done with a concise single line of code.

As nice as it is, getting the hang of the streaming operator is tough.  The SystemVerilog 1800-2012 LRM isn't totally clear (at least to me) on the details of how they work.  The statement from the LRM that really got me was this, "The stream_operator << or >> determines the order in which blocks of data are streamed: >> causes blocks of data to be streamed in left-to-right order, while << causes blocks of data to be streamed in right-to-left order."  You might have some intuitive idea about which end of a stream of bits is on the "right" and which is on the "left" but, I sure didn't.  After looking at the examples of streaming on page 240 of the LRM I thought I had it, and then none of my attempts to write streaming concatenations worked like I thought they should.  Here's why: "right" and "left" are different depending on whether your stream of bits is a packed array or an unpacked array.

As far as I can tell, "right" and "left" are in reference to the literal SystemVerilog code representations of an arrays of bits.  A literal packed array is generally written like this:

bit [7:0] packed_array = 8'b0011_0101;

And packed_array[0] is on the right (that 1 right before the semicolon).  A literal unpacked array is written like this:

bit unpacked_array[] = '{1'b1, 1'b0, 1'b1, 1'b0};

unpacked_array[0] is on the left (the first value after the left curly brace).  I don't know about you, but I'm generally more concerned with actual bit positions, not what is to the right and left in a textual representation of an array, but there you have it.

Once I got that down, I still had problems.  It turns out the results of streaming concatenations will be different depending on the variable you are storing them in.  It's really the same right/left definitions coming into play.  If you are streaming using the right-to-left (<<) operator, the right-most bit of the source will end up in the left-most bit of the destination.  If your destination is a packed array then, just as I explained above, "right" means bit zero and left means the highest index bit.  If, your destination is an unpacked array, your right-most source bit will end up as bit zero of the unpacked array (which is the "right" bit according to the literal representation).

Got all that?  If not, I put a code example on edaplayground that you can run and examine the output of.  The examples are all streaming bits one at a time.  It gets a little harder to wrap your head around what happens when you stream using a slice_size and when your source and/or destinations array is an unpacked array of bytes or ints. I'll write another post explaining some tricks for those next (UPDATE: next post is here).