Summary:
Add an argument to the Exit instruction. Update the LLVM semantics to
execute the Exit instruction and store the result in an "exited"
component of the state. (Previously it just noticed that it was stuck
about to do an Exit.)
With exiting treated uniformly, now in the proof that for every LLVM
trace, there is a llair trace that simulates it, all of the cheats
except for 1 are just cases that I haven't got to yet. However, the last
cheat is for the situation where the LLVM program gets stuck and the
llair program doesn't. For example, the following two line LLVM program
gets stuck because r2 is not assigned (ignoring for the moment the static
restriction that LLVM is in SSA form).
r1 := r2
Exit(0)
The compilation to llair omits the assignment and so we get a llair
program that doesn't get stuck:
Exit(0)
The key question is whether the static restrictions are sufficient to
ensure that no expression that might be omitted can get stuck.
Reviewed By: jberdine
Differential Revision: D17737589
fbshipit-source-id: bc6c01a1b
Summary:
If the LLVM to llair translation keeps a mapping from register r to
expression e, then for each register r' mentioned in e, there must be an
assignment to r' that dominates the entire live range of r. Thus, where
ever r might be replaced by e, the value of r' will be the same as it
was when the initial assignment to r occurred. Maintaining this
invariant relies on the LLVM being in SSA form.
Reviewed By: jberdine
Differential Revision: D17710288
fbshipit-source-id: fd3eaa57d
Summary:
Since the correcteness of the mapping from LLVM to llair depends on
LLVM being SSA, we need to formalise what that means. We also prove that
the domination relation is a strict partial order, which will probably
be helpful when reasoning about the translation.
Reviewed By: jberdine
Differential Revision: D17631456
fbshipit-source-id: a00eb3f87
Scott Owens