@graphql-codegen 创建的 `Exact<T>` 类型的目的是什么？

Question

GraphQL Code Generator 在创建的 TypeScript 文件的顶部创建此类型：

export type Exact<T extends { [key: string]: unknown }> = { [K in keyof T]: T[K] };

并将其用于所有客户端创建的查询变量：

src/foo.graphql:

query Foo($id: ID!) {
  foo(id: $id) {
    bar
  }
}

generated/foo.ts:

...

export type Exact<T extends { [key: string]: unknown }> = { [K in keyof T]: T[K] };

...

export type FooQueryVariables = Exact<{
  id: Scalars['ID'];
}>;

...

这种 Exact<T> 类型的目的是什么？它如何影响 FooQueryVariables（相对于如果它不存在）？

https://www.graphql-code-generator.com/#live-demo的完整演示：

schema.graphql:

schema {
  query: Query
}

type Query {
  foo(id: ID!): Foo
}

type Foo {
  bar: String!
}

operation.graphql:

query Foo($id: ID!) {
  foo(id: $id) {
    bar
  }
}

codegen.yml:

generates:
  operations-types.ts:
    plugins:
      - typescript
      - typescript-operations

生成operations-types.ts:

export type Maybe<T> = T | null;
export type Exact<T extends { [key: string]: unknown }> = { [K in keyof T]: T[K] };
export type MakeOptional<T, K extends keyof T> = Omit<T, K> & { [SubKey in K]?: Maybe<T[SubKey]> };
export type MakeMaybe<T, K extends keyof T> = Omit<T, K> & { [SubKey in K]: Maybe<T[SubKey]> };
/** All built-in and custom scalars, mapped to their actual values */
export type Scalars = {
  ID: string;
  String: string;
  Boolean: boolean;
  Int: number;
  Float: number;
};

export type Query = {
  __typename?: 'Query';
  foo?: Maybe<Foo>;
};


export type QueryFooArgs = {
  id: Scalars['ID'];
};

export type Foo = {
  __typename?: 'Foo';
  bar: Scalars['String'];
};

export type FooQueryVariables = Exact<{
  id: Scalars['ID'];
}>;


export type FooQuery = { __typename?: 'Query', foo?: Maybe<{ __typename?: 'Foo', bar: string }> };

Answer 1

它旨在使人们无法将具有任何附加属性（除 id 之外）的对象传递为 FooQueryVariables。但它没有这样做：https://github.com/dotansimha/graphql-code-generator/issues/4577

@graphql-codegen 创建的 `Exact<T>` 类型的目的是什么？

What is the purpose of the `Exact<T>` type that @graphql-codegen creates?

typescript

graphql

graphql-codegen