英文:
Conditional type check for undefined
问题
以下是代码的翻译部分：
type stringUndefined = "string" | undefined;
type What<T> = T extends undefined ? "true" : "false";
const no: What<stringUndefined> = "";
no 变成了 "true" | "false"，而不是我所期望的 "true"。
英文:
For the code
type stringUndefined = "string" | undefined;
type What<T> = T extends undefined ? "true" : "false";
const no : What<stringUndefined> = "";
no becomes "true" | "false" instead of what I would expect, "true"
Edit:
strict null checks are enabled
答案1
得分: 3
"string" | undefined 不扩展 undefined，因为它可以是 "string"。
但 undefined 扩展 "string" | undefined，因为联合类型的成员扩展（细化）联合类型。所以：
type StringLiteralOrUndefined = "string" | undefined;
type What<T> = undefined extends T ? true : false;
type X = What<StringLiteralOrUndefined>;
// ^? type X = true
type UnrelatedType = string | number;
type Y = What<UnrelatedType>;
// ^? type Y = false
英文:
"string" | undefined doesn't extend undefined, because it can be "string".
But undefined extends "string" | undefined, because the members of a union extend (refine) the union. So:
type StringLiteralOrUndefined = "string" | undefined;
type What<T> = undefined extends T ? true : false;
type X = What<StringLiteralOrUndefined>;
// ^? type X = true
type UnrelatedType = string | number;
type Y = What<UnrelatedType>;
// ^? type Y = false
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