类型“{}”不能赋值给类型“T”。

Question

类型“{}”不能赋值给类型“T”。

11

function f1<T extends {[prop: string]: any}}>(a: T = {}) {
    console.log(a)
}

错误：类型'{}'不能赋值给类型'T'。

我试图给一个默认值，而不是给T一个默认类型。

我不确定问题出在哪里。

- Stav Alfi

1个回答

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- Fyodor Yemelyanenko · Accepted Answer

考虑以下例子：

type A = { [prop: string]: any };
type B = { [prop: string]: any; prop1: string };

类型B是否扩展类型A？是的！

但以下内容将不起作用。

let a: A = {};
let b: B = a; // Error

很明显，类型 A 缺少必要的属性 prop1，所以变量 a 无法赋值给变量 b。

你的函数也存在相同的问题。当你执行以下操作时

function f1<T extends { [prop: string]: any }>(a: T = {}) {
    console.log(a);
}

编译器说对象{}不能分配给类型T。如果这样做可以的话，你可以这样做：

f1<B>();  // a will equal to {} and misses required prop1

这可能看起来并不像错误，因为在 f1 中你只知道 T extends { [prop: string]: any } ，并且对 prop1 一无所知。但是请考虑如果你想返回 T：

function f1<T extends { [prop: string]: any }>(a: T = {}): T {
    console.log(a);
    return a;
}

如果这段代码能够正常工作的话，它将会引入一个bug。

let bb: B = f1<B>();  // Here bb will be equal to empty object
let s: string = bb.prop1;  // Compiler will not show error, as it thinks that bb has prop1 of type string. But in reality it is undefined.

所以有以下解决方案可供考虑：

Remove all generics. Type { [prop: string]: any } is generic by itself so it may suit your needs

function f1(a: { [prop: string]: any } = {}) {
    console.log(a);
    return a;
}

Make a fully optional. In this case a may equal to undefined and compiler would know about it.
```
function f1<T extends { [prop: string]: any } = {}>(a?: T) {
    console.log(a);
}
```
If you return a compiler will tell you that you should either return union type (with undefined) or check in function body that a is not undefined.
Worst case from my point of view is to use type cast as already suggested
```
function f1<T extends { [prop: string]: any }>(a: T = {} as T) {
    console.log(a);
}
```
But be careful and don't forget that you can miss some required properties in this case.