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[英]TypeScript: How to make generic type to infer inside a function?
[英]How to make Typescript infer a generic method argument type in an object
我怎样才能使这项工作?
const variant1 = (data: { hello: number }) => {/* Do some work here... */};
const variant2 = (data: { hi: string }) => {/* Do some work here... */};
interface Data<T = unknown> {
callback: (data: T) => void;
}
const f: Record<string, Data> = {
name1: variant1, // '(data: {hello: number}) => void' is not assignable to type 'Data<unknown>'
name2: variant2, // '(data: {hi: string}) => void' is not assignable to type 'Data<unknown>'
}
/*
What I want:
const f: Record<string, Data> = {
name1: variant1, // (data: { hello: number }) => void
name2: variant2, // (data: { hi: string }) => void
}
*/
换句话说,我希望 Typescript 插入方法参数的实际类型而不是泛型。
我们可以在不从f
中删除显式类型的情况下实现吗?
您可以从f
变量中删除类型定义,TypeScript 将自动解析类型
const variant1 = (data: { hello: number }) => {/* Do some work here... */};
const variant2 = (data: { hi: string }) => {/* Do some work here... */};
const f = {
name1: variant1,
name2: variant2,
}
f.name1({ hello: 1 });
f.name2({ hi: "" });
这是游乐场的链接
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