Question

I m translating a piece of code from javascript to coffeescript.

  for (var i = 0, len = keys.length; i < len; i++) {
    k = keys[i];
    if (!mailOptions.hasOwnProperty(k))
      mailOptions[k] = app.set( mailOptions )[k]
  }

我不知道如何对待这一问题,在专题网站上也不清楚,有人能否给我解释? 非常感谢!

Answer 1

for key in keys
  if not mailOptions.hasOwnProperty key
    mailOptions[key] = (app.set  mailOptions )[key]

Or guard-style:

for key in keys when not mailOptions.hasOwnProperty key
  mailOptions[key] = (app.set  mailOptions )[key]

:

var key, _i, _len;

for (_i = 0, _len = keys.length; _i < _len; _i++) {
  key = keys[_i];
  if (!mailOptions.hasOwnProperty(key)) {
    mailOptions[key] = (app.set( mailOptions ))[key];
  }
}

Answer 2

http://js2coffee.org/“rel=“nofollow”http://js2coffee.org/:

i = 0
len = keys.length

while i < len
  k = keys[i]
  mailOptions[k] = app.set("mailOptions")[k]  unless mailOptions.hasOwnProperty(k)
  i++

But I wouldn t do it this way. I would just do:

for k in keys
  mailOptions[k] = app.set("mailOptions")[k] unless mailOptions.hasOwnProperty k

This outputs the following (excluding var, which it also outputs):

for (_i = 0, _len = keys.length; _i < _len; _i++) {
  k = keys[_i];
  if (!mailOptions.hasOwnProperty(k)) {
    mailOptions[k] = app.set("mailOptions")[k];
  }
}

Or, if you wanted to be fancier, which I don t advise in this situation, since it sacrifices some readability:

(mailOptions[k] = app.set("mailOptions")[k] unless mailOptions.hasOwnProperty k) for k in keys

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