如果接口具有泛型参数，如何从类型定义(`.d.ts`)导入接口

Question

下面是提出这个问题时gulp-intercept的类型定义：

/// <reference types="node" />

import Vinyl = require('vinyl');

declare namespace intercept {
    interface Intercept {
        (interceptFunction: InterceptFunction): NodeJS.ReadWriteStream;
    }

    interface InterceptFunction {
        (file: Vinyl): Vinyl;
    }
}

declare var intercept: intercept.Intercept;

export = intercept;

Vynil的Typings允许定义自定义属性。因此，如果我们在intercept函数中编写类似file.customProperty = 1的代码，就不会出现TypeScript错误。然而，对于自动完成，我想扩展Vynyl接口并重写类型，如：

import Vinyl = require('vinyl');

declare namespace intercept {
    interface Intercept<VinylFile__PossiblyWithCustomProperties extends Vinyl> {
        (interceptFunction: InterceptFunction<VinylFile__PossiblyWithCustomProperties>): NodeJS.ReadWriteStream;
    }

    interface InterceptFunction<VinylFile__PossiblyWithCustomProperties extends Vinyl> {
        (file: VinylFile__PossiblyWithCustomProperties): VinylFile__PossiblyWithCustomProperties;
    }
}

declare var intercept: intercept.Intercept;

export = intercept;

行declare var intercept: intercept.Intercept中有错误

TS2314: Generic type `VinylFile__PossiblyWithCustomProperties` requires 1 argument(s).

在这里，我们不知道将使用哪个Vynil超集，因此我不确定declare var intercept: intercept.Intercept<Vynil>；是否正确。

Answer 1

我不确定我是否完全理解了这种情况，但如果您只是想编译这个示例，请在Intercept的函数级别注释类型参数VinylFile__PossiblyWithCustomProperties (在下面重命名为T )，而不是作为顶级接口声明的一部分：

import Vinyl = require("vinyl");

declare namespace intercept {
  interface Intercept {
    // annotate type parameter T directly at function
    <T extends Vinyl>(interceptFunction: InterceptFunction<T>): NodeJS.ReadWriteStream;
  }

  interface InterceptFunction<T extends Vinyl> {
    (file: T): T;
  }
}

declare var intercept: intercept.Intercept;

export = intercept;

客户端中的调用示例：

type ExtendedFile = Vinyl & { foo: string };

declare const callback: (file: ExtendedFile) => ExtendedFile;

intercept(callback);