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Only hash dep node indices of deps of anon tasks #60573

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Merged
merged 1 commit into from
May 7, 2019
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Only hash dep node indices of deps of anon tasks #60573

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21 changes: 9 additions & 12 deletions src/librustc/dep_graph/graph.rs
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Original file line number Diff line number Diff line change
Expand Up @@ -1021,25 +1021,22 @@ impl CurrentDepGraph {
fn complete_anon_task(&mut self, kind: DepKind, task_deps: TaskDeps) -> DepNodeIndex {
debug_assert!(!kind.is_eval_always());

let mut fingerprint = self.anon_id_seed;
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Can you add a comment explaining why this works? Especially why it is OK to stable-hash unstable data like DepNodeIndex values?

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Done.

let mut hasher = StableHasher::new();

for &read in task_deps.reads.iter() {
let read_dep_node = self.data[read].node;
// The dep node indices are hashed here instead of hashing the dep nodes of the
// dependencies. These indices may refer to different nodes per session, but this isn't
// a problem here because we that ensure the final dep node hash is per session only by
// combining it with the per session random number `anon_id_seed`. This hash only need
// to map the dependencies to a single value on a per session basis.
task_deps.reads.hash(&mut hasher);

::std::mem::discriminant(&read_dep_node.kind).hash(&mut hasher);
let target_dep_node = DepNode {
kind,

// Fingerprint::combine() is faster than sending Fingerprint
// through the StableHasher (at least as long as StableHasher
// is so slow).
fingerprint = fingerprint.combine(read_dep_node.hash);
}

fingerprint = fingerprint.combine(hasher.finish());

let target_dep_node = DepNode {
kind,
hash: fingerprint,
hash: self.anon_id_seed.combine(hasher.finish()),
};

self.intern_node(target_dep_node, task_deps.reads, Fingerprint::ZERO).0
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