问题描述
接口定义如下:
interface IRemoteService {
createRecord(record: RecordType): ng.IPromise<ICreateResponse<string>>;
}
interface ICreateResponse<T> {
createdId: T;
}
以下代码为什么不引起Typescript编译错误?
Why doesn't the following code cause a Typescript compilation error?
class RemoteServiceMock implements IRemoteService {
public static $inject = ["$q"];
constructor(private $q: ng.IQService){
}
createRecord(record: RecordType): ng.IPromise<ICreateResponse<string>> {
return this.$q.when({});
}
}
$q.when
的类型是when<T>(value: T): IPromise<T>
.
推荐答案
这是根据规范.这是简化的示例:
This is according to spec. Here is your example simplified:
interface A{
}
interface B {
createdId: string;
}
var foo:ng.IPromise<A>;
var bar:ng.IPromise<B>;
bar = foo; // No error
如果A是B的子类型或B是A的子类型,则允许此分配.如果不是这种情况,则会出现如下所示的错误:
This assignment is allowed if A is a subtype of B or B is a subtype of A. If this is not the case you will get an error as shown below:
interface A {
breakTypeCompat: number;
}
interface B {
createdId: string;
}
var foo:ng.IPromise<A>;
var bar:ng.IPromise<B>;
bar = foo; // Error
原因是函数参数的双方兼容性.请参阅此链接以获取文档+这是为什么的原因: https://github.com/Microsoft/TypeScript/wiki/Type-Compatibility#function-argument-bivariance
The reason is the bivariance compatibility of function arguments. See this link for docs + reason why this is the way it is: https://github.com/Microsoft/TypeScript/wiki/Type-Compatibility#function-argument-bivariance
接口的类型兼容性取决于您如何使用它们.例如.以下不是错误:
Type compatibility of the interfaces depends upon how you use them. E.g. the following is not an error :
interface IPromise<T>{
}
interface A{
}
interface B {
createdId: string;
}
var foo:IPromise<A>;
var bar:IPromise<B>;
bar = foo; // No error
但是,如果IPromise
在哪里将type参数用作成员,则会出错:
However if the IPromise
where to use the type parameter as a member it would error:
interface IPromise<T>{
member:T
}
interface A{
}
interface B {
createdId: string;
}
var foo:IPromise<A>;
var bar:IPromise<B>;
bar = foo; // Error
因此
在实际的承诺定义中,我们有类似以下内容:
Therefore
In the actual promise definition we have something like:
interface IPromise<T> {
then(successCallback: (promiseValue: T) => any): any;
}
interface A {
}
interface B {
createdId: string;
}
var foo: IPromise<A>;
var bar: IPromise<B>;
bar = foo; // No Error
由于我们将T
用作函数的参数,因此A
和B
将通过双方差检查类型.因此,如果A是B的子集或B是A的子集,则它们是兼容的.
Since we are using T
as an argument to the a function A
and B
will be type checked by bivariance. So if A is a subset of B or B is a subset of A they are compatible.
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