`[source] <https://gitlab.haskell.org/ghc/ghc/tree/master/compiler/cmm/CmmExpr.hs>`_

compiler/cmm/CmmExpr.hs
=======================


Note [Old Area]
~~~~~~~~~~~~~~~

`[note link] <https://gitlab.haskell.org/ghc/ghc/tree/master/compiler/cmm/CmmExpr.hs#L89>`__

There is a single call area 'Old', allocated at the extreme old
end of the stack frame (ie just younger than the return address)
which holds:
  * incoming (overflow) parameters,
  * outgoing (overflow) parameter to tail calls,
  * outgoing (overflow) result values
  * the update frame (if any)

Its size is the max of all these requirements.  On entry, the stack
pointer will point to the youngest incoming parameter, which is not
necessarily at the young end of the Old area.

End of note 



Note [CmmStackSlot aliasing]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`[note link] <https://gitlab.haskell.org/ghc/ghc/tree/master/compiler/cmm/CmmExpr.hs#L106>`__

When do two CmmStackSlots alias?

 - T[old+N] aliases with U[young(L)+M] for all T, U, L, N and M
 - T[old+N] aliases with U[old+M] only if the areas actually overlap

Or more informally, different Areas may overlap with each other.

An alternative semantics, that we previously had, was that different
Areas do not overlap.  The problem that lead to redefining the
semantics of stack areas is described below.

e.g. if we had

::

    x = Sp[old + 8]
    y = Sp[old + 16]

..

    Sp[young(L) + 8]  = L
    Sp[young(L) + 16] = y
    Sp[young(L) + 24] = x
    call f() returns to L

if areas semantically do not overlap, then we might optimise this to

    Sp[young(L) + 8]  = L
    Sp[young(L) + 16] = Sp[old + 8]
    Sp[young(L) + 24] = Sp[old + 16]
    call f() returns to L

and now young(L) cannot be allocated at the same place as old, and we
are doomed to use more stack.

  - old+8  conflicts with young(L)+8
  - old+16 conflicts with young(L)+16 and young(L)+8

so young(L)+8 == old+24 and we get

    Sp[-8]  = L
    Sp[-16] = Sp[8]
    Sp[-24] = Sp[0]
    Sp -= 24
    call f() returns to L

However, if areas are defined to be "possibly overlapping" in the
semantics, then we cannot commute any loads/stores of old with
young(L), and we will be able to re-use both old+8 and old+16 for
young(L).

::

    x = Sp[8]
    y = Sp[0]

..

    Sp[8] = L
    Sp[0] = y
    Sp[-8] = x
    Sp = Sp - 8
    call f() returns to L

Now, the assignments of y go away,

    x = Sp[8]
    Sp[8] = L
    Sp[-8] = x
    Sp = Sp - 8
    call f() returns to L