箭头样式用户定义的类型保护？

Question

typescript handbook on user defined type guards将示例类型保护定义为

function isFish(pet: Fish | Bird): pet is Fish {
    return (<Fish>pet).swim !== undefined;
}

箭头功能是否有相应的语法？

Answer 1

使用类型转换而不是类型声明：

const isFish = (pet => !!pet.swim) as (pet) => pet is Fish 

然而，鉴于它更冗长，我宁愿将类型保护写为普通函数，除非你真的需要this绑定，但这可能是代码味道。

Answer 2

是的，就像您返回boolean一样，只需返回pet is Fish：

const isFish: (pet: Fish | Bird) => pet is Fish = pet => (pet as Fish).swim !== undefined

箭头函数被声明为单个类型（(pet: Fish | Bird) => pet is Fish）而不是参数和返回类型。

Answer 3

TypeScript支持箭头功能类型防护（2017年可能不是这种情况）。现在，以下各项将按预期工作。我在生产代码中经常使用这种样式：

const isFish = (pet: Fish | Bird): pet is Fish => 
  (pet as Fish).swim !== undefined;

Answer 4

作为替代方案，这也有效

const isFish = (pet: any): pet is Fish => !!pet.swim;

或者如果您希望更明确

const isFish = (pet: any): pet is Fish => pet.swim === 'function';

同时可以对属性进行测试

const isFish = (pet: any): pet is Fish => pet.hasOwnProperty('genus');