Comments (2)
Your bottom examples "work" because CommonUtilFns<Common>
and CommonUtilFns<A | B>
is not the same as CommonUtilFns<A> | CommonUtilFns<B>
. You can't just distribute across unions. CommonUtilFns<T>
is invariant in T
:
declare let f: CommonUtilFns<A> | CommonUtilFns<B>
declare let g: CommonUtilFns<A | B>
f = g; // error!
g = f; // error!
Your top example doesn't work because it's essentially #30581. TS does not keep track of the identity of util
here, just its type. If you have util1
and util2
both of type CommonUtilFns<A> | CommonUtilFns<B>
, then if (util1.isAorB(val)) return util2.getProp(val)
wouldn't be safe, right? That's the problem TS has. And that's been reported in #30581 and the recommended approach is detailed at #47109. That looks like the following:
interface UtilMap {
A: A;
B: B;
}
const utils: { [K in keyof UtilMap]: CommonUtilFns<UtilMap[K]> } = {
A: {
isAorB: (version: Common): version is A => !!(version as A).propA,
getProp: (version: A) => ({ prop: version.propA }),
},
B: {
isAorB: (version: Common): version is B => !!(version as B).propB,
getProp: (version: B) => ({ prop: version.propB }),
}
};
function apply<K extends keyof UtilMap>(util: (typeof utils)[K], val: Common) {
return util.isAorB(val) ? util.getProp(val) : undefined
}
{
const getProp = (val: Common) => {
for (const util of Object.values(utils)) {
const ret = apply(util, val);
if (ret) return ret;
}
return null;
};
}
Which, yes, is a lot of weird generic stuff. See #47109. Anyway I'd say this is essentially a duplicate of #30581.
from typescript.
Thanks for the detailed explanation @jcalz! Your apply
function was also more or less what I got recommended by Copilot, which is to do both the guarding and the prop extraction in a single function, with the same generic context. It's an understandable limitation, but a little frustrating because TS gets so close to getting it right.
I'm going to close this issue. Maintainers, feel free to mark it as a duplicate. And thanks again for your time Joe 👍
from typescript.
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