TypeScript 索引类型对象和记录之间的断开推理

uwopmtnx 于 8个月前发布在 TypeScript

关注(0)|答案(4)|浏览(248)

TypeScript版本： 2.2.2
代码

interface Errors {
  [field: string]: {
    key: string
    message: string
  }
}
const errors: Errors = {}
function genericObjectFunction<K extends string, V>(obj: Record<K, V>): [K, V][] {
  return []
}
/*
Argument of type 'Errors' is not assignable to parameter of type 'Record<string, never>'.
Index signatures are incompatible.
Type '{ key: string; message: string; }' is not assignable to type 'never'.
*/
genericObjectFunction(errors)

尽管类型看起来大致兼容，但这是合法的：

const errorRecord: Record<string, { key: string, message: string }> = errors

预期行为：

genericObjectFunction(errors) 应该能够编译。

typescript

来源：https://github.com/microsoft/TypeScript/issues/14930

4条答案

按热度按时间

4ngedf3f1#

这个失败的原因是

interface Errors {
  [field: string]: {
    key: string
    message: string
  };
}

没有离散的键，所以记录是空的。 Record<K, V> 类型是通过将键Map到值来定义的，由于没有已知的值，通用示例化没有成员。
考虑

const errors = { // no explicit type, inferred type has members, still compatible with interface.
  password: {
    key: 'minimumComplexity',
    message: 'Password must contain at least one non alphanumeric character'
  }
};
function genericObjectFunction<K extends string, V>(obj: Record<K, V>): [K, V][] {
  return [];
};
const genericErrors = genericObjectFunction(errors);

编辑：

这里有一个带有解决方法的例子(注意：重载顺序很重要)

interface Errors {
  [field: string]: {
    key: string
    message: string
  }
}
const errors: Errors = {
  password: {
    key: 'minimumComplexity',
    message: 'Password must contain at least one non alphanumeric character'
  }
};
function genericObjectFunction<K extends string, V>(obj: Record<K, V>): [K, V][];
function genericObjectFunction<V>(obj: { [field: string]: V }): [string,{[P in keyof V]:V[P]}][];
function genericObjectFunction<K extends string, V>(obj: Record<string, V>): [K, V][]{
  return [];
};
const typeInferredErrors = {
  password: {
    key: 'minimumComplexity',
    message: 'Password must contain at least one non alphanumeric character'
  }
};
const genericErrorsX = genericObjectFunction(errors);
const genericErrorsY = genericObjectFunction(typeInferredErrors);