使用拉姆达咖喱+；带有TypeScript的占位符结果为；类型'；占位符'；不可分配给类型'；x'；

我想将已知参数绑定到一个带有Ramda的回调，其余参数未指定：

import * as R from "ramda";

type Params = Partial<{
  param1: number;
  param2: number;
  param3: number;
}>;

function getBoundCallback(params: Params) {
  const { param1, param2, param3 } = params;

  const callback = (p1: number, p2: number, p3: number) => {
    // do something
  };

  return R.curry(callback)(
    param1 || R.__, 
    param2 || R.__, 
    param3 || R.__
  );
}

正确的方法是什么

我能想到的唯一解决办法是：

const callback = (
    _p1: number | R.Placeholder,
    _p2: number | R.Placeholder,
    _p3: number | R.Placeholder
  ) => {
    const p1 = _p1 as number;
    const p2 = _p2 as number;
    const p3 = _p3 as number;

    // do something
  };

这远远不是最优的

正确的方法是什么

我使用

“types/npm-ramda#dist”

进行ramda类型声明

const callback = (
    _p1: number | R.Placeholder,
    _p2: number | R.Placeholder,
    _p3: number | R.Placeholder
  ) => {
    const p1 = _p1 as number;
    const p2 = _p2 as number;
    const p3 = _p3 as number;

    // do something
  };