microsoft / microsoft/TypeScript
Different behavior when declare property or method in a class about covariance
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描述
TypeScript Version: 4.1.2
Search Terms: In ts doc - type compatibility, it seems like not mentioned about this.
Code
class M1 {
c: number = 1;
}
export class M2 extends M1 {
a: number = 1;
}
type Func<M extends M1> = (m: M) => number;
interface IFunc<M extends M1> {
func1?: Func<M>,
func2: Func<M>,
}
class R1<M extends M1 = M1> implements IFunc<M> {
// when delete this property, it will be no error.
public func1 = (m: M) => {
return m.c
}
public func2(m: M) {
return m.c
};
}
type IRConstructor<M extends M1, R extends R1<M>> = new () => R;
export class R2 extends R1<M2> {
}
type A = IRConstructor<M1, R1>;
let a1: A = R2; // error: Type 'typeof R2' is not assignable to type 'IRConstructor<M1, R1<M1>>'.
console.log(a1);
Expected behavior:
Should declare a property or a method in class has same behavior? I'm not sure...
Actually I want to know how can I make these code without error when I have to use functional property.
Actual behavior:
When you just decalre a method which is compatible with IFunc<M1>, it will be fine. But when you declare a functional property which is compatible with IFunc<M1>, it will be an error.
Playground Link:
Or, you can see playground
Related Issues:
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调研方向
从 Issue 中链接的 TypeScript Playground 复现和 Issue 中引用的 handbook「Type Compatibility」部分开始。比较 generic class 在 functional property 形式和 method 形式下报告的行为,然后确定这种差异是否是有意的;完成的要求是确定预期行为,并在 compiler 或 documentation 中提供相应的覆盖。
由索引模型根据 Issue 内容生成。
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- 缺陷
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