Ignored error in class constructor causes method usage errors to be ignored

[katz12]

TypeScript Version: 3.5.3

Search Terms:
ts-ignore in class constructor
class method no error on parameter usage

Code

class Foo {
  bar: string
  constructor(bar: string) {
    this.bar = bar
  }

  baz(buzz: number) {
    return buzz + 1
  }
}

// @ts-ignore
const instance = new Foo()
instance.baz('buzz')

Expected behavior:
The invocation of instance.baz('buzz') should report that the parameter should be a number instead of a string.

Actual behavior:
There is no error on that line.
If I fix the ts-ignore on the line above, it then reports the error correctly.

Playground Link:
http://www.typescriptlang.org/play/?ts=3.8.0-dev.20200116#code/MYGwhgzhAEBiD29oG8BQ1oCMwCcBc0EALjgJYB2A5utMPOcTgK7BHw4AU2+hJFlAShQ0MRABakIAOm7QAvFlw0AvqhrYAXlyYaNBckwC2mAKY4haDBhwmiTHOSw6N0ANTQAjCtSrUAej9oAAEiCABaUkpydhNUOgYiaApiMHJgE3lochMAdzhEDgFUZKJU9JkwLQByTGcqgSA

Related Issues:
Could not find any through Google or here.

[DanielRosenwasser]

This is confusing because from just hovering over instance, the type is Foo; however, I think what's happening is that we have different type-checker modes, and new Foo() produces something different from each type-checker.

new Foo() in our error-producing type-checker returns the internal error type that works like any because it tries to avoid cascading errors on subsequent operations. However, operations like quick info and auto-completerely on a more lax type-checker that tries to get results of operations in a best-effort way and produces the type Foo. However, that more relaxed checker isn't the same type-checker instance that produces errors, so you don't get an error on that subsequent operation.

@sheetalkamat @weswigham does that sound about right?

[weswigham]

That is precisely it. See #28564 or #28950 for a possible fix.

[katz12]

Thanks for the responses. I will focus on fixing type errors in constructors for now.