使用流指定内联函数的返回类型

Question

我有以下功能：

const safeNull = fn => (txt: string): string => (isNil(txt) ? '' : fn(txt));

export const stripSpaces: Function = safeNull(txt => txt.replace(/\s/g, ''));

export const safeTrim: Function = safeNull(txt => txt.trim());

stripSpaces 和 safeTrim 返回字符串怎么说。

Answer 1

您的safeNull 函数被键入以返回一个返回字符串的函数。 所以你所要做的就是从stripSpaces 和safeTrim 中删除那些Function 类型。 Flow 将推断它们返回字符串，因为 safeNull 返回类型。

const safeNull = fn => (txt: string): string => (isNil(txt) ? '' : fn(txt));

export const stripSpaces = safeNull(txt => txt.replace(/\s/g, ''));

export const safeTrim = safeNull(txt => txt.trim());

如果您愿意，也可以显式定义它们的类型，如下所示：

const safeNull = fn => (txt: string): string => (isNil(txt) ? '' : fn(txt));

export const stripSpaces: string => string = safeNull(txt => txt.replace(/\s/g, ''));

export const safeTrim: string => string = safeNull(txt => txt.trim());