使用flow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());

如何说stripSpaces和safeTrim返回字符串。

Answer 1

键入您的safeNull函数以返回返回字符串的函数。 因此，您只需从Function和stripSpaces中删除safeTrim种类型。 由于safeNull返回类型，Flow会推断它们返回字符串。

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，也可以明确定义他们的类型，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());