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3ae7ab698f
Save liveness results for DestinationPropagation `DestinationPropagation` needs to verify that merge candidates do not conflict with each other. This is done by verifying that a local is not live when its counterpart is written to. To get the liveness information, the pass runs `MaybeLiveLocals` dataflow analysis repeatedly, once for each propagation round. This is quite costly, and the main driver for the perf impact on `ucd` and `diesel`. (See https://github.com/rust-lang/rust/pull/115105#issuecomment-1689205908) In order to mitigate this cost, this PR proposes to save the result of the analysis into a `SparseIntervalMatrix`, and mirror merges of locals into that matrix: `liveness(destination) := liveness(destination) union liveness(source)`. <details> <summary>Proof</summary> We denote by `'` all the quantities of the transformed program. Let $\varphi$ be a mapping of locals, which maps `source` to `destination`, and is identity otherwise. The exact liveness set after a statement is $out'(statement)$, and the proposed liveness set is $\varphi(out(statement))$. Consider a statement. Suppose that the output state verifies $out' \subset phi(out)$. We want to prove that $in' \subset \varphi(in)$ where $in = (out - kill) \cup gen$, and conclude by induction. We have 2 cases: either that statement is kept with locals renumbered by $\varphi$, or it is a tautological assignment and it removed. 1. If the statement is kept: the gen-set and the kill-set of $statement' = \varphi(statement)$ are $gen' = \varphi(gen)$ and $kill' = \varphi(kill)$ exactly. From soundness requirement 3, $\varphi(in)$ is disjoint from $\varphi(kill)$. This implies that $\varphi(out - kill)$ is disjoint from $\varphi(kill)$, and so $\varphi(out - kill) = \varphi(out) - \varphi(kill)$. Then $\varphi(in) = (\varphi(out) - \varphi(kill)) \cup \varphi(gen) = (\varphi(out) - kill') \cup gen'$. We can conclude that $out' \subset \varphi(out) \implies in' \subset \varphi(in)$. 2. If the statement is removed. As $\varphi(statement)$ is a tautological assignment, we know that $\varphi(gen) = \varphi(kill) = \\{ destination \\}$, while $gen' = kill' = \emptyset$. So $\varphi(in) = \varphi(out) \cup \\{ destination \\}$. Then $in' = out' \subset out \subset \varphi(in)$. By recursion, we can conclude by that $in' \subset \varphi(in)$ everywhere. </details> This approximate liveness results is only suboptimal if there are locals that fully disappear from the CFG due to an assignment cycle. These cases are quite unlikely, so we do not bother with them. This change allows to reduce the perf impact of DestinationPropagation by half on diesel and ucd (https://github.com/rust-lang/rust/pull/115105#issuecomment-1694701904). cc ````@JakobDegen```` |
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