Tasks flamegraph generation #3048
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| const children = toFold.children.map((c) => foldFramegraph(c)); | ||
| children.sort((a, b) => (a.parallel === b.parallel ? 0 : 1)); |
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This looks suspicious to me. It seems like the intention is to put all the parallel items together, but I'm pretty sure that's not what this code will do.
My other problem is that whatever order is set here will be broken by the next sort.
What would make sense to me would be to put all the parallel items at the beginning, and the non-parallel at the end, and to then sort each half by name:
children.sort((a, b) => {
if (a.parallel === b.parallel) {
return a.name.localeCompare(b.name);
}
return a.parallel ? -1 : 1;
});But I don't know if this is what you had in mind.
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The intended behavior is similar to what you proposed. But take into account that Array.prototype.sort is a stable sort.
We first sort by parallel, then by name. So that all of the children with the same name are together, and within a name-group, the parallel ones are also together.
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Please ping me on slack if you want to go over this together.
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I understand this now. What I didn't understand was that the goal is to have the items grouped by name, and sorted internally by parallel, not the other way around (sorted by parallel state, and internally by name).
We need to fix the sort by parallel to return a -1 too though.
Co-authored-by: Franco Victorio <victorio.franco@gmail.com>
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First PR of the compilation times optimization effort.
This PR introduces a Hardhat argument
--flamegraph, which creates a flamegraph of the Hardhat tasks.