Question

假设：

interface B {
    base: string;
}
interface C1 extends B {
    c1: string;
}
interface C2 extends B {
    c2: string;
}

type A = C1 | C2;

var aOk1: A = {
    base: '',
    c1: '',
}
var aOk2: A = {
    base: '',
    c2: '',
}
var a: A = {
    base: '',
    c1: '', // was expecting this to error at compile time
    c2: '', // was expecting this to error at compile time
}

a.c1; // Correctly errors
a.c2; // Correctly errors

正如代码中标记的那样，我期望分配c1和c2属性导致编译时错误。有没有办法实现这个目标？

为了澄清动机，这是因为设置选项对象有两种相互排斥的方式，我希望使用以下方法在定义（.d.ts）文件中键入它： / p>

// definition in a .d.ts file
someFunction(options: C1 | C2)

因此，如果有人试图错误地传递两个值设置的选项对象，它们将在编译时显示错误，但是他们当前可以执行以下操作而不会出现任何编译时错误：

// consuming in a .ts file

// User might think both options will be used but
// actually c2 option silently overrides c1.
someFunction({base: '', c1: '', c2: ''});

** 编辑：标记的联合类型 **

您无法使用tagged union types也称为"Discriminated Unions"解决此问题：

interface C1 extends B {
    kind: 'C1',
    c1: string;
}
interface C2 extends B {
    kind: 'C2',
    c2: string;
}

type A = C1 | C2;

var a: A = {
    kind: 'C1',
    base: '',
    c1: '', // would like this to compile time error
    c2: '', // would like this to compile time error
};

Answer 1

似乎没有解决方案，但这里有相关的讨论：

https://github.com/Microsoft/TypeScript/issues/14094 https://github.com/Microsoft/TypeScript/issues/12997 https://github.com/Microsoft/TypeScript/issues/12745

用于限制赋值的Typescript类型

1 个答案: