[TS] Type narrowing string union?

I have the following setup;

const fn = (source: keyof typeof Obj | "$none") => {
  if (source in Obj) Obj[source];
  //                 ^^^^^^^^^^^
  // Element implicitly has an 'any' type because expression of type [...] can't be used to index type [...]
  // more code...
}

const fn = (source: keyof typeof Obj | "$none") => {
  if (source in Obj) Obj[source];
  //                 ^^^^^^^^^^^
  // Element implicitly has an 'any' type because expression of type [...] can't be used to index type [...]
  // more code...
}

I also have the following relevant compiler options;

"strict": true,
"alwaysStrict": true,
"keyofStringsOnly": true,
"moduleResolution": "node",
"module": "esnext",
"target": "esnext"

"strict": true,
"alwaysStrict": true,
"keyofStringsOnly": true,
"moduleResolution": "node",
"module": "esnext",
"target": "esnext"

Is there a way that I can prevent the error from occuring and have typescript correctly narrow source's type from keyof typeof Obj | "$none" to keyof typeof Obj? I would have thought that if (source in Obj) then obviously typeof source === keyof typeof Obj, if not directly then due to Obj's keys definitely including source. I can't figure out how to narrow it, I've tried .startsWith("$") but for some reason that's still not a type predicate to a ${arg0}${string} literal, Object.keys(Obj).includes(source) is just an ugly equivalent to the codeblock above, source[0] === "$" doesn't work. Nothing except Obj[source as keyof typeof Obj] seems to work, which is not only ugly but entirely defeats the purpose of using typescript there.

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[TS] Type narrowing string union?