Typescript 使用强制转换编写打字函数的惯用方法

您问题中方法的问题在于，您试图根据仅在编译时可用的类型脚本类型信息来控制JavaScript运行时行为.

JavaScript运行时只看到engineProperties object，并不知道您的任何类型脚本接口或类型，这就是instanceof或typeof不适用于这种情况的原因.您可以考虑以下几种方法:

1.使用类而不是接口或类型

JavaScript支持classes，因此TypeScript类信息将在编译后继续存在，并可用于JavaScript运行时.这意味着您可以使用instanceof操作符:

class PetrolEngineProperties {
    constructor(
        public cylinders: number,
        public capacity: number
    ) {}
}

class ElectricEngineProperties {
    constructor(
        public numberOfBatteries: number,
        public batteryCellSize: number
    ) {}
}

type EngineProperties = PetrolEngineProperties | ElectricEngineProperties;

function initializeCarEngine(engineProperties: EngineProperties) {
    if (engineProperties instanceof ElectricEngineProperties) {
        console.log('Electric engine was started');
    } else if (engineProperties instanceof PetrolEngineProperties) {
        console.log('Petrol engine was started');
    }
}

initializeCarEngine(new ElectricEngineProperties(2, 6));

2.使用区别对待的联合

将判别式属性添加到您的类型，并创建一个discriminated union:

interface PetrolEngineProperties {
    kind: 'PETROL';
    cylinders: number;
    capacity: number;
}

interface ElectricEngineProperties {
    kind: 'ELECTRIC';
    numberOfBatteries: number;
    batteryCellSize: number;
}

type EngineProperties = PetrolEngineProperties | ElectricEngineProperties;

function initializeCarEngine(engineProperties: EngineProperties) {
    if (engineProperties.kind === 'ELECTRIC') {
        console.log('Electric engine was started');
    } else if (engineProperties.kind === 'PETROL') {
        console.log('Petrol engine was started');
    }
}

initializeCarEngine({
    kind: 'PETROL',
    cylinders: 8,
    capacity: 2400
});

3.写类型谓词

如果您发现自己处于不能或不想向对象添加属性的情况，则可以创建type predicates以使TypeScrip编译器相信对象是基于其现有内容的特定类型:

interface PetrolEngineProperties {
    cylinders: number;
    capacity: number;
}

interface ElectricEngineProperties {
    numberOfBatteries: number;
    batteryCellSize: number;
}

type EngineProperties = PetrolEngineProperties | ElectricEngineProperties;

function initializeCarEngine(engineProperties: EngineProperties) {
    if (isElectric(engineProperties)) {
        console.log('Electric engine was started');
    } else if (isPetrol(engineProperties)) {
        console.log('Petrol engine was started');
    }
}

function isElectric(engineProperties: EngineProperties)
        : engineProperties is ElectricEngineProperties {
    return 'numberOfBatteries' in engineProperties
        && 'batteryCellSize' in engineProperties;
}

function isPetrol(engineProperties: EngineProperties)
        : engineProperties is PetrolEngineProperties {
    return 'cylinders' in engineProperties
        && 'capacity' in engineProperties;
}

initializeCarEngine({
    cylinders: 8,
    capacity: 2400
});

还有其他可能的解决方案，但要记住的主要一点是，在TypeScrip中，大多数类型信息在编译时消失，您不能将其用作运行时决策的基础.