What is the best way to handle errors in a class method?

Question

Node.js

Hello. I have a class with two methods. At the moment, I handle errors like this:

class Geo {
   async method() {
      try {
          return await _method()
      } catch (err) {
          throw err
      }
   }

   async _method() {
      try {
         return await someFunc()
      } catch (err) {
         throw err
      }
   }
}

app.post('/test', async (req, res) => {
  try {
     const geo = new Geo()
     geo.method()
  } catch (err) {
     console.log(err)
  }
})

Do I understand correctly that I can remove try/catch from Geo class methods? After all, if I wrap a call to geo.method () in try / catch, then everything will work the same way. Otherwise, I don’t want to fence redundant try / catch everywhere

Answer 1

1. Design:

try {
  // что угодно
} catch (err) {
  throw err;
}

doesn't make sense. Because it is equivalent to:
2. Using an asynchronous function with only one return:

async method() {
  return await _method();
}

just as pointless and the code can be simplified to:

method() {
  return  _method();
}

3. "fencing redundant try/catch everywhere" is wrong because:

const foo = () => {
  try {
    bar();
  } catch (e) {
    // этот блок не будет вызван никогда, так как ошибка перехватывается в вызове bar()
  }
};

const bar = () => {
  try {
    dangerousCall();
  } catch (e) {
    // этот блок будет вызван в случае ошибки вызова dangerousCall()
  }
}

try {
  foo();
} catch (e) {
  // этот блок не будет вызван никогда, так как ошибка перехватывается в вызове bar()
}

Demo
4. Handle the error where the function that can trigger it is executed.
No, that would be the wrong decision. See point 4.