JavaScript Node - Private Module Functions

Question

To learn the language of JavaScript, I did a dojo exercise by Roy Osherove. Is there a better way to organize this so that the only method exported is add while keeping it in one single module?

module.exports = {
  add: function (string) {
    this._checkForError(string);
    this._checkForNegatives(string);
    return this._result(string);
  },

  _result: function (string) {
    var numbers = this._numbers(string);
    var result = numbers.reduce(function (sum, number) {
      return sum + number;
    });
    return result;
  },

  _numbers: function (string) {
    var modded = this._modded(string);
    var numbers = modded.map(function (number) {
      return parseInt(number, 10) || 0;
    });
    return numbers;
  },

  _checkForError: function (string) {
    if (string.match(/\n$/)) {
      throw 'Expression ending in newline!';
    }
  },

  _negatives: function (string) {
    var numbers = this._numbers (string);
    var negatives = numbers.filter(function (number) {
      return number < 0;
    });
    return negatives;
  },

  _delimiter: function (string) {
    return (string.substr(0, 2) === '//') ? string[2] : ',';
  },

  _checkForNegatives: function (string) {
    var negatives = this._negatives(string);
    if (negatives.length > 0) {
      throw 'Negatives not allowed: ' + negatives.join(', ');
    }
  },

  _modded: function (string) {
    var delimiter = this._delimiter(string);
    return string.replace(/\n/, delimiter)
      .split(delimiter);
  }
};

I see what you're saying, would you mind illustrating a simple example? @Dagg — user27606, Jun 2 '14 at 3:27
Just use a regular function declaration that isn't assigned to a property of the exports object. — Dagg, Jun 2 '14 at 3:28
After spending all night reading on modules, I realized that the old module pattern does the exact same thing as module.exports. It took me awhile to get my head around that. In this case, I can wrap the module in a function and at the end attach a return to the function add() and define the remaining methods above to exclude it from being "exposed". @Dagg — user27606, Jun 2 '14 at 15:20

nrw · Accepted Answer · 2014-06-02 17:26:01Z

module.exports is just an object. Set properties on it that you want to expose.

exports.add = add

This is also equivalent in this case.

module.exports = {add: add}

Now, when you require this module, you'll get an object with one property: add.

There are a few other small issues:

Nothing about your design depends on accessing methods via this. This module is clearer without bringing this into the mix.

function add (string) {
  _checkForError(string)
  _checkForNegatives(string)
  return _result(string)
}

This code is quite easy to follow already, but It would be even better if you had your methods in the order they are called so the code reads like an article: broad info at the top, specific details later on.

I'd also leave out semicolons where they will be handled by automatic semicolon insertion, but that's a style choice.

function _checkForError (string) {
  if (string.match(/\n$/)) {
    throw 'Expression ending in newline!'
  }
}

function _checkForNegatives (string) {
  var negatives = _negatives(string)
  if (negatives.length > 0) {
    throw 'Negatives not allowed: ' + negatives.join(', ')
  }
}

function _negatives (string) {
  var numbers = _numbers(string)
  var negatives = numbers.filter(function (number) {
    return number < 0
  })
  return negatives
}

function _result (string) {
  var numbers = _numbers(string)
  var result = numbers.reduce(function (sum, number) {
    return sum + number
  })
  return result
}

function _numbers (string) {
  var modded = _modded(string)
  var numbers = modded.map(function (number) {
    return parseInt(number, 10) || 0
  })
  return numbers
}

function _modded (string) {
  var delimiter = _delimiter(string)
  return string.replace(/\n/, delimiter).split(delimiter)
}

function _delimiter (string) {
  return (string.substr(0, 2) === '//') ? string[2] : ','
}

Thanks for explaining this, I appreciate it. – user27606 Jun 2 '14 at 21:42 — user27606, Jun 2 '14 at 21:42

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