javascript - typescript 编译器类型赋值错误被忽略

Question

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。这似乎是因为 Customer和 CustomerLayoutViewModel完全相同。

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

所以我的问题是:如何配置编译器以在上面的示例中产生错误？

Answer 1

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有额外的属性类型仍然兼容，没有机会有人会通过 customerViewModel 访问不存在的东西:

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 会出现兼容性错误具有额外的必需属性:

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(private)字段添加到类中。如果名称相同，私有(private)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'.