箭头样式用户定义的类型保护?

TypeScript支持箭头函数类型guards(2017年可能没有)。现在,以下各项按预期工作。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;

使用类型断言而不是类型声明:

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

然而,考虑到它更为冗长,我更愿意将类型保护作为普通函数编写,除非您确实需要 this 绑定,但这可能是代码的味道。

是的,就像你回来时一样 boolean ,您只需返回 pet is Fish :

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

箭头函数声明为单一类型( (pet: Fish | Bird) => pet is Fish )而不是将参数和返回类型分开。

作为一种替代方案,这也可行

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

或者如果你想更明确一点

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

而测试也可以针对属性进行

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