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literal function with structure spread can't be rightly inferred in function's generic type slot #28816
Description
TypeScript Version: 3.2.1
Search Terms:
strucutre spread + generic
Code
function id<T>(input: T): T { return input }
function getFoo ({ foo = 42 }) {
return foo
}
const newGetFoo = id(getFoo)
const newGetFoo2 = id(function getFoo ({ foo = 42 }) {
return foo
})Expected behavior:
newGetFoo and newGetFoo2 should have the same signature
Actual behavior:
newGetFoo is:
({ foo }: {
foo?: number | undefined;
}) => numberbut newGetFoo2 is
({ foo }: any) => numberRelated Issues:
Activity
ahejlsberg commentedon Jan 25, 2019
Our intended design is to determine the type of a parameter from one of the following, in order of preference:
See original discussion and rationale here.
For the
getFoofunction we end up using the type implied by the binding pattern (i.e.({ foo }: { foo?: number }) => number) because there is no type annotation, contextual type, or initializer type. This is working as intended.However, for the function expression passed as an argument to
idwe do have a contextual type, namelyT. SinceThas no constraint we should end up inferring an implicitanytype for the parameter, but this isn't quite working correctly.So, the bug here is that
newGetFoo2should have type({ foo }: any) => anyand in-noImplicitAnymode we should report an implicit any error onfoo.EDIT: In retrospect, I now agree
newGetFoo2should have the same type asgetFooandnewGetFoo. #36476 has the fix.ahejlsberg commentedon Jan 28, 2020
Fix in #36476 now produces the result expected in the original post.