代码之家 › 专栏 › 技术社区 › Kyung Lee

为什么在传递给条件类型时,联合字符串泛型类型被视为特定的文字?

conditional-types union-types typescript-generics typescript

1

Kyung Lee · 技术社区 · 7 月前

打字版本:5.6.2

为什么 ApprovalType 被篡改为特定值 批准类型 如果函数参数类型是条件类型,即使我显式传递泛型参数 <ApprovalType> 调用函数时?

type ApprovalType = "PENDING" | "APPROVED" | "REJECTED";

type A<T> = {
    options: T[];
};

type B<T> = {
    options: T[];
};

function conditionalTypeFn<T>(props: T extends string ? B<T> : A<T>) {
    return props;
}
conditionalTypeFn<ApprovalType>({
    /**
     * â Type '("PENDING" | "APPROVED" | "REJECTED")[]' is not assignable
     * to type '"PENDING"[] | "APPROVED"[] | "REJECTED"[]'
     */
    options: ["PENDING", "APPROVED", "REJECTED"],
});

function unionTypeFn<T>(props: A<T> | B<T>) {
    return props;
}
unionTypeFn<ApprovalType>({
    /* â no error */
    options: ["PENDING", "APPROVED", "REJECTED"],
});

1 回复 | 直到 7 月前

1

Robby Cornelissen 7 月前

默认情况下,如中所述 documentation ,条件类型在 分配的 在联合类型上使用时的方式。

这里有一个简化的例子来证明这一点:

type ApprovalType = "PENDING" | "APPROVED" | "REJECTED";

type Distributed<T> = T extends string ? T[] : never;

type ApprovalTypes = Distributed<ApprovalType>;
// type ApprovalTypes = "PENDING"[] | "APPROVED"[] | "REJECTED"[]

为了避免分布式行为,您可以在条件类型中使用方括号。应用于同一个例子,这将产生:

type ApprovalType = "PENDING" | "APPROVED" | "REJECTED";

type NonDistributed<T> = [T] extends [string] ? T[] : never;

type ApprovalTypes = NonDistributed<ApprovalType>;
// type ApprovalTypes = ("PENDING" | "APPROVED" | "REJECTED")[]

或者,应用于您问题中的代码:

function conditionalTypeFn<T>(props: [T] extends [string] ? B<T> : A<T>) {
    return props;
}

^{Playground link}