如何修复一个javascript async fn包装一个跳到底部返回值的.then回调函数？，也使用'react-native-permissions'

问：

我的异步函数看起来像

import { PERMISSIONS, RESULTS, checkMultiple } from 'react-native-permissions';
import { Platform } from 'react-native';
  const checkPermissions = async (): boolean => {
    console.log('Platform.OS', Platform.OS)
    if (Platform.OS === 'ios') {
      checkMultiple([
        PERMISSIONS.IOS.LOCATION_ALWAYS,
        PERMISSIONS.IOS.MOTION,
      ]).then((status) => {
        if (
          status[PERMISSIONS.IOS.LOCATION_ALWAYS] === RESULTS.GRANTED &&
          status[PERMISSIONS.IOS.MOTION] === RESULTS.GRANTED
        ) {
          console.log('permissions allgood');
          return true;
        } else {
          console.log('PERMISSIONS are false.');
          return false;
        }
      });
    } else if (Platform.OS === 'android') {
      checkMultiple([
        PERMISSIONS.ANDROID.ACCESS_BACKGROUND_LOCATION,
        PERMISSIONS.ANDROID.ACTIVITY_RECOGNITION,
      ]).then((status) => {
        if (
          String(status[PERMISSIONS.ANDROID.ACCESS_BACKGROUND_LOCATION]) ===
            RESULTS.GRANTED &&
          String(status[PERMISSIONS.ANDROID.ACTIVITY_RECOGNITION]) ===
            RESULTS.GRANTED
        ) {
          console.log('permissions allgood');
          return true;
        } else {
          console.log('PERMISSIONS are false.');
          return false;
        }
      });
    }
    return 'why';
  };

    const permissionsOn = checkPermissions();
    console.log('permissionsOn',permissionsOn)

结果是： LOG Platform.OS ios LOG permissionsAreON before {“_h”： 0， “_i”： 1， “_j”： “why”， “_k”： null}//'why' LOG 权限 allgood

那么如何格式化包装 fn.then（） 的异步函数并返回结果呢？