这必须是一种简单的方法:
const gfn = <T>(a: T) => a
// Doesn't work
const cloned = gfn<number>
如何在已设置泛型的情况下获取克隆函数?
答案 0 :(得分:0)
看起来没有一种简单的方法来分配type参数。
两个选项是:
const cloned: (a: number) => number = gfn; // this does not clone anything
const cloned2 = (a: number) => gfn(a); // actually a new function