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typescript中可为null的函数变量的正确类型是什么?

typescript-typings typescript

0

Anthony Kong · 技术社区 · 6 年前

我有这段代码

let anchorRenderer: string => React.Component;  // line 658
if (link) {
    if (userAnchorRenderer) {
      anchorRenderer = userAnchorRenderer;  // line 661
    } else {
      anchorRenderer = (link: string) => <a className={styles.link} href={link}/>    // line 663
    }
}

...

// Usage of the function 
return <div>
    {anchorRenderer && anchorRenderer(link)})   
 </div>

但是我不能正确地得到类型声明。

在这个版本中,我得到了以下错误:

Error:(663, 11) TS2322: Type 'P["anchorRenderer"]' is not assignable to type 'string'.
  Type '((link: string) => Component<{}, {}, any>) | undefined' is not assignable to type 'string'.
    Type 'undefined' is not assignable to type 'string'.
Error:(663, 11) TS2322: Type '(link: string) => Element' is not assignable to type 'string'.

如果我把变量声明改为

let anchorRenderer: string => React.Component | undefined;

Error:(658, 35) TS2362: The left-hand side of an arithmetic operation must be of type 'any', 'number' or an enum type.

1 回复 | 直到 6 年前

1

Nurbol Alpysbayev 6 年前

改变

let anchorRenderer: string => React.Component | undefined;

到

let anchorRenderer: (str: string) => React.Component | undefined;

https://github.com/Microsoft/TypeScript/issues/3693 ),还必须在类型声明中使用paren(与js arrow函数相反)。