I'm trying to figure out why the compiler doesn't infer A as a union type like string[] | number[] when it fails. Instead, A is inferred as the first return value, which in this case is string[]. Is there a solution to this issue?
const define = <A>({ handler }: { handler: (some: boolean) => A[] }) => ({ handler });
const a = define({
handler: (some: boolean) => {
if (some) {
return ["foo", "bar"];
} else {
return [1, 2];
}
},
});
This issue arises due to the behavior of Typescript when analyzing your code. Specifically, when it encounters the following statement -
if (some) {
return ["foo", "bar"];
}
Typescript eagerly attempts to infer the type for the generic because it is a plain generic without any constraints. As a result, it expects the type to be consistent with the values returned. If the subsequent code returns an array of numbers instead of strings, Typescript goes back to verify the function definition and raises an error.
To address this issue, you can make use of the following approach -
const define = <A extends Array<unknown>,>({ handler }: { handler: (some: boolean) => A }) => ({ handler });
const a = define({
handler: (some: boolean) => {
if (some) {
return ["foo", "bar"];
} else {
return [1, 2];
}
},
});
You can experiment further with this solution using this Playground link.
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