Conditional type narrowing for function arguments doesn't work

[vshab]

TypeScript Version: 3.4.0-dev.201xxxxx

Search Terms:
conditional type narrowing function arguments

Code

type IfNull<T, U> = T extends null ? U : undefined;

interface Value {
  do: () => void;
}

// foo type shows as "function foo(error: Error | null, value: Value | undefined): void"
function foo(error: Error | null, value: IfNull<typeof error, Value>) {
  if (error) {
    return;
  }
  
  // error TS2532: Object is possibly 'undefined'.
  value.do();
}

// Shouldn't be allowed:
foo(null, undefined);
foo(new Error(), { do: () => {} });

// However these work as expected:
let error: Error | null = null,
  value: IfNull<typeof error, Value> = { do: () => {} };

value.do();

strictNullChecks is true.

Expected behavior:
Typescript should narrow types for function arguments as it does for variable declaration.

Actual behavior:
Function arguments are treated without narrowing.

Playground Link:
https://www.typescriptlang.org/play/#src=type%20IfNull%3CT%2C%20U%3E%20%3D%20T%20extends%20null%20%3F%20U%20%3A%20undefined%3B%0D%0A%0D%0Ainterface%20Value%20%7B%0D%0A%20%20do%3A%20()%20%3D%3E%20void%3B%0D%0A%7D%0D%0A%0D%0A%2F%2F%20foo%20type%20shows%20as%20%22function%20foo(error%3A%20Error%20%7C%20null%2C%20value%3A%20Value%20%7C%20undefined)%3A%20void%22%0D%0Afunction%20foo(error%3A%20Error%20%7C%20null%2C%20value%3A%20IfNull%3Ctypeof%20error%2C%20Value%3E)%20%7B%0D%0A%20%20if%20(error)%20%7B%0D%0A%20%20%20%20return%3B%0D%0A%20%20%7D%0D%0A%20%20%0D%0A%20%20%2F%2F%20error%20TS2532%3A%20Object%20is%20possibly%20'undefined'.%0D%0A%20%20value.do()%3B%0D%0A%7D%0D%0A%0D%0A%2F%2F%20Shouldn't%20be%20allowed%3A%0D%0Afoo(null%2C%20undefined)%3B%0D%0Afoo(new%20Error()%2C%20%7B%20do%3A%20()%20%3D%3E%20%7B%7D%20%7D)%3B%0D%0A%0D%0A%2F%2F%20However%20these%20work%20as%20expected%3A%0D%0Alet%20error%3A%20Error%20%7C%20null%20%3D%20null%2C%0D%0A%20%20value%3A%20IfNull%3Ctypeof%20error%2C%20Value%3E%20%3D%20%7B%20do%3A%20()%20%3D%3E%20%7B%7D%20%7D%3B%0D%0A%0D%0Avalue.do()%3B

Related Issues:
Didn't find.

[jack-williams]

A typeof type used in a signature will simply pick out the declared type of the identifier; there is no fundamental link back to the argument that will evolve over time. As you identify: if you look at the type of foo presented by the checker you see that any notion of error has been erased.

function foo(error: Error | null, value: Value | undefined): void { … }

One adjustment that will correctly flag the calls of foo that you would like to error is:

function foo<T extends Error | null>(error: Error | T, value: IfNull<T, Value>) { … }

To narrow value based on error is non-trivial, essentially it is dependent typing.

If you really want narrowing in the function you might have to do something like this:

function foo(...args: [Error, null] | [undefined, Value]) {
  if (args[0]) {
    return;
  }
  args[1].do();
}

[vshab]

@jack-williams Thank you for clarification.
But isn't it wrong? I mean why is the function signature being changed and as you said error has been erased?

One adjustment that will correctly flag the calls of foo that you would like to error is:

Interesting... Thanks! But I'm really more interested in narrowing inside of the function.

If you really want narrowing in the function you might have to do something like this:

Great! However when I'm trying to make it more user-friendly i.e.:

function foo(...[error, value]: [Error, null] | [undefined, Value]) {
  if (error) {
    return;
  }

  value.do();
}

It doesn't work. Am I missing something?

[jack-williams]

Type narrowing does not work across variables. This is a design limitation that is tracked here #12184.

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.