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Adding infer in a conditional type changes the resulting branch #22615
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Anders Hejlsberg (@ahejlsberg) bug indeed?
It is effectively a design limitation. We have the concept of instantiating a generic function type in the context of a non-generic function type (a form of unification), but we currently don't do that in conditional types. Instead, function type parameters are erased to their constraints and we infer from those. In the example above, that causes us to infer
{}for theinfer Rtype parameter. Ordinarily that would be fine because anything is assignable to{}, however in this case the type parameter occurs in a contra-variant position, so the check fails following inference. This isn't specific to conditional types, a similar thing can happen during regular type inference in function calls.The proper fix for this would be have type inference perform instantiation of a source type in the context of a target type when the source type is a generic function type.
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on Jul 25, 2018
TypeScript Version: 2.8.0-dev.20180315
Search Terms:
infer conditional generic function
Code
The code checks whether a function that takes the identity function is assignable to a function that takes some function
any => any. This is clearly assignable, and thus results in "yes". However, when replacing oneanybyinfer R, the other branch is taken. I think that replacinganyby aninfer Rtype should never change the branch that is taken.{ "compilerOptions": { "allowJs": true, "target": "es6", "module": "commonjs", "outDir": "dest", "strictNullChecks": true, "jsx": "preserve", "strictFunctionTypes": true } }Expected behavior:
XandYresolve to"yes"Actual behavior:
Yresolves to"no"Related Issues: