[LLVMdev] New backend help request.

[James Boulton]

That's very helpful, thank-you!

I've seen immediate offsets handled through an Operand define. How on earth does it know to put an offset into simm16 below along with the address in the register? It seems to be a more implicit way of doing the immediate add pattern you suggested, but I have no idea how this actually works.

def simm16 : Operand<i32>;

def mem : Operand<i32> 
{
	let MIOperandInfo = (ops AR, simm16);
	let PrintMethod = "printMemOperand";
}

I prefer the explicit pattern method suggested, as I can understand what's going on, but I would really like to know how the Operand stuff actually works.

Thanks!


-----Original Message-----
From: Tim Northover [mailto:t.p.northover at gmail.com] 
Sent: 09 July 2015 19:15
To: James Boulton
Cc: LLVM Dev
Subject: Re: [LLVMdev] New backend help request.

> As that just turns into stores, sets, etc. But how would you represent things like indexed access?
>
> move dn, (an,dn)
> move dn, offset(an)

These are fairly easy (the difficulty with the pre/post indexed modes
is that they tend to define two values or otherwise not be tree-like
in DAG form).

For these you just put in a more complicated pattern representing the
address calculation that happens. For example, with "(an, dn)" the
address is "an+dn" so you'd write your pattern as: "(store i32:$Dn,
(add i32:$An, i32:$Dm)". For the "Dn.W" variants you'll probably want
to throw a trunc in there.

And for the offset ones you'll want to define an "offsetimm" (or
whatever your preferred name is) instance of ImmLeaf with i32 type
(because addresses are 32-bits) but only accepting signed 16-bit
values that can actually be encoded:

def offsetimm : ImmLeaf<i32, [{ return Imm >= -32768 && Imm <= 32767; }]>;
[...]
(store i32:$Dn, (add i32:$An, offsetimm:$imm))

The biggest wrinkle is that you probably want to defer choosing
whether to use An or Dn as the offset as long as possible. You could
either fix that up later (optimisation!) or try to define these
instructions to accept *both* Dn and An and then encode it properly.
I'm afraid I don't know enough about the 68k instruction set to be
sure which is better.

Cheers.

Tim.