infer-from-usage suggestions can't be ignored, and always do something when invoked. - #28206
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Nathan Shively-Sanders (sandersn) merged 4 commits intoOct 29, 2018
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Still need to write tests.
Nathan Shively-Sanders (sandersn)
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Daniel Rosenwasser (DanielRosenwasser)
October 29, 2018 17:34
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Note that we should do this for unused locals and unused parameters errors as well, since they also can't be ts-ignored, but offer it nonetheless. |
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Oct 29, 2018
| const type = inferTypeForVariableFromUsage(setAccessorDeclaration.name, program, cancellationToken) || | ||
| inferTypeForVariableFromUsage(param.name, program, cancellationToken); | ||
| let type = inferTypeForVariableFromUsage(setAccessorDeclaration.name, program, cancellationToken); | ||
| if (type === program.getTypeChecker().getAnyType()) { |
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Would it be better to use type.flags & TypeFlags.Any here?
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Short answer: no.
Long answer: here we are checking for exactly the fallback type produced inside inferFromUsage.ts, not other types that come from the checker like errorType. Those types are already filtered out in addCandidateType, which does use TypeFlags.Any.
Nathan Shively-Sanders (sandersn)
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October 29, 2018 20:23
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The ts-ignore codefix does not apply to suggestions, so I made separate implicit any suggestions.
Based on our discussion with VS Code, and with Brian Terlson (@bterlson). I agree that invoking a suggestion and having nothing happen feels like a bug, which is worse than inferring a useless
any.