[Stack Switching] Improve cont.bind fuzzing#8500
[Stack Switching] Improve cont.bind fuzzing#8500kripken wants to merge 18 commits intoWebAssembly:mainfrom
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kripken
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Mar 19, 2026
kripken
commented
- Pick a signature from the known interesting types.
- Pick a continuation type using that signature.
src/tools/fuzzing/fuzzing.cpp
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| auto newCont = Continuation(newSig); | ||
| auto newType = Type(newCont, NonNullable, Exact); | ||
| std::vector<Expression*> newArgs{make(newParam)}; | ||
| auto numOldParams = sig.params.size(); |
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Shouldn't this be numNewParams, since it's the number of parameters of the new continuation cont.bind creates? Maybe "input" and "output" would be clearer than "new" and "old."
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Sounds good, renamed to input/output.
src/tools/fuzzing/fuzzing.cpp
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| if (pickedSig.results != sig.results) { | ||
| // Results must match. | ||
| continue; | ||
| } |
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WDYT about filtering out impossible signatures before picking? We could memoize the results to avoid pathological repeated work.
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Good idea, and probably fast enough without memoizing. Done.
src/tools/fuzzing/fuzzing.cpp
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| auto numAddedParams = numNewParams - numOldParams; | ||
| bool bad = false; | ||
| for (Index i = 0; i < numOldParams; i++) { | ||
| if (pickedSig.params[numAddedParams + i] != sig.params[i]) { |
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This should allow for subtyping in one direction or the other. The params of the output signature should be subtypes of the corresponding params of the input signature.
src/tools/fuzzing/fuzzing.cpp
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| if (numinputParams < numOutputParams) { | ||
| // Too short. | ||
| continue; | ||
| } | ||
| // Ignoring the input params at the start, compare the tails. | ||
| auto numAddedParams = numinputParams - numOutputParams; | ||
| bool bad = false; | ||
| for (Index i = 0; i < numOutputParams; i++) { | ||
| if (!Type::isSubType(outputSig.params[i], | ||
| pickedSig.params[numAddedParams + i])) { | ||
| bad = true; | ||
| break; | ||
| } | ||
| } |
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This filtering can be moved up, too. Then we won't need to try to find an appropriate signature multiple times at all.
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