忽略打字稿编译器类型分配错误



我有以下情况:

const customer = new Customer();
let customerViewModel = new CustomerLayoutViewModel();
customerViewModel = customer;

这在编译时不会产生错误,在我看来这是错误的行为。这似乎是由于CustomerCustomerLayoutViewModel完全相同的事实。

问题是随着时间的推移,它们将不相同,我希望上面的代码给出编译错误,因为类型不同。

所以我的问题:如何配置编译器以在上面的例子中产生错误?

Typescript 在确定类型兼容性时使用结构类型。这意味着,如果这两种类型具有兼容的结构,它们将是兼容的:

class Customer { name: string }
class CustomerLayoutViewModel { name: string }
const customer = new Customer();
let customerViewModel = new CustomerLayoutViewModel();
customerViewModel = customer; // OK compatible

如果Customer具有额外的属性,则类型仍然兼容,则某人不可能通过不存在customerViewModel某些内容进行访问:

class Customer { name: string; fullName: string }
class CustomerLayoutViewModel { name: string }
const customer = new Customer();
let customerViewModel = new CustomerLayoutViewModel();
customerViewModel = customer; // still OK compatible

如果CustomerLayoutViewModel具有额外的必需属性,则会出现兼容性错误:

class Customer { name: string }
class CustomerLayoutViewModel { name: string; fullName: string }
const customer = new Customer();
let customerViewModel = new CustomerLayoutViewModel();
customerViewModel = customer; //error now

确保类型不兼容的一种方法是将私有字段添加到类中。 如果名称相同,则私有字段将与任何其他类事件中的任何其他字段不兼容:

class Customer { private x: string }
class CustomerLayoutViewModel { private x: string }
const customer = new Customer();
let customerViewModel = new CustomerLayoutViewModel();
customerViewModel = customer; //error Types have separate declarations of a private property 'x'. 

这不会给出编译错误,因为您没有分配customercustomerViewModel的类型,但是如果您执行以下操作,那么您应该得到编译时错误:

const customer:Customer = new Customer();
let customerViewModel:CustomerLayoutViewModel = new CustomerLayoutViewModel();
customerViewModel  = customer;

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