Infer extra parameter types from default values #17559

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1. olegdunkan commented on Aug 2, 2017

   

   olegdunkan

   on Aug 2, 2017

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   When a function expression with no type parameters and no parameter type annotations is contextually typed (section 4.23) by a type T and a contextual signature S can be extracted from T, the function expression is processed as if it had explicitly specified parameter type annotations as they exist in S. Parameters are matched by position and need not have matching names. If the function expression has fewer parameters than S, the additional parameters in S are ignored. If the function expression has more parameters than S, the additional parameters are all considered to have type Any.

   Isn't that case?
2. DanielRosenwasser commented on Aug 2, 2017

   

   Daniel Rosenwasser (DanielRosenwasser)

   on Aug 2, 2017

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   Thanks for linking to that olegdunkan! It's indeed the specified behavior, though it's certainly surprising and it looks like noImplicitAny doesn't catch it.
3. Daniel Rosenwasser (DanielRosenwasser)

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   on Aug 2, 2017
4. falsandtru commented on Aug 2, 2017

   

   falsandtru (falsandtru)

   on Aug 2, 2017

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   This spec seems to say about default behaviors that has no default value. Anders Hejlsberg (@ahejlsberg) Does this spec mean that type inference has to ignore a type of default value too and still should be so even today?
5. simonbuchan commented on Oct 8, 2017

   

   Simon Buchan (simonbuchan)

   on Oct 8, 2017

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   Seems like a spec bug: considered to have type Any should probably be whatever spec language would be for considered to have the type that would be inferred without a contextual signature.
6. Ryan Cavanaugh (RyanCavanaugh)

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   Needs More InfoThe issue still hasn't been fully clarifiedThe issue still hasn't been fully clarified

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7. RyanCavanaugh commented on Oct 10, 2017

   

   Ryan Cavanaugh (RyanCavanaugh)

   on Oct 10, 2017

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   falsandtru (@falsandtru) how did you even end up in this state?
8. falsandtru commented on Oct 10, 2017

   

   falsandtru (falsandtru)

   on Oct 10, 2017

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   To define and initialize local variables. For example,

   const count = ((a = 0) => () => ++a)();

   can be used as shorthand for

   const count = (() => {
       let a = 0;
       return () => ++a;
   })();
9. Ryan Cavanaugh (RyanCavanaugh)

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   on Oct 10, 2017
10. falsandtru commented on Oct 31, 2017

    

    falsandtru (falsandtru)

    on Oct 31, 2017

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    Sorry, my previous example is wrong. Here is a correct one.

    f((a = 0) => ++a);
    
    function f(cb: () => void) {
    }
11. RyanCavanaugh commented on Jun 23, 2021

    

    Ryan Cavanaugh (RyanCavanaugh)

    on Jun 23, 2021

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    Example in OP now works as hoped
12. Ryan Cavanaugh (RyanCavanaugh)

    closed this as completedon Jun 23, 2021
13. simonbuchan commented on Jun 24, 2021

    

    Simon Buchan (simonbuchan)

    on Jun 24, 2021

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    Aww, I was hoping that that would mean this works:

    function getOr<T, K extends keyof T, D>(
      obj: T,
      key: K,
      // Type 'undefined' is not assignable to type 'D'.
      //  'D' could be instantiated with an arbitrary type which could be unrelated to 'undefined'.(2322)
      defaultValue: D = undefined,
    ): T[K] | D {
        if (key in obj) {
            return obj[key];
        } else {
            return defaultValue;
        }
    }
    
    // $ExpectType string | undefined
    getOr(process.env, "SOME_VAR");
    // $ExpectType string | null
    getOr(process.env, "SOME_VAR", null);
    // $ExpectType string
    getOr(process.env, "SOME_VAR", "DEFAULT");