我正在尝试编写一个memoize函数,它将函数作为参数并返回一个类似的memoized函数。
function memoize<T extends Function, R>(f: T): T {
const memory = new Map<string, R>();
const g = (...args: any[]) => {
if (!memory.get(args.join())) { memory.set(args.join(), f(...args)); }
return memory.get(args.join());
};
return g; // g as T => [ts] Type '(...args: any[]) => R' cannot be converted to type 'T'.
}
// const exp: (...args: any[]) => RegExp
const exp = memoize<(text: string) => RegExp, RegExp>((text: string) => {
return new RegExp(text.replace(/[^a-zA-Z0-9\s]/g, ".").replace(/\s+/g, "\\s+"), "ig");
});
问题在于,如果我只返回g
,则exp的签名变为(...args: any[]) => RexExp
,如果我尝试强制g为T,则ts
会抱怨g不可分配到T
。
有没有办法“强迫”g
成为同一类型的f
,以便exp
将完全相同的函数类型传递给memoize
}?
答案 0 :(得分:4)
这似乎有效:
function memoize<R, T extends (...args: any[]) => R>(f: T): T {
const memory = new Map<string, R>();
const g = (...args: any[]) => {
if (!memory.get(args.join())) {
memory.set(args.join(), f(...args));
}
return memory.get(args.join());
};
return g as T;
}
const exp = memoize((text: string) => {
return new RegExp(text.replace(/[^a-zA-Z0-9\s]/g, ".").replace(/\s+/g, "\\s+"), "ig");
});