CODEWITHSUNDEEP
coffeescript
trebuchet
trebuchet

Reputation: 1481

for...in if not null or undefined

In my Javascript code, I process many json objects with properties that may be null:

if (store.departments != null) {
    for(var i = 0; i < store.departments.length; i++) {
        alert(department.name);
    }
}

In porting my app to coffeescript, I came up with the following shortcut using the existential operator:

for department in store.departments ? []
    alert department.name

Is this acceptable coffeescript? Is there any scenario in which this would not work as intended?

Upvotes: 2

Views: 2430

Answers (2)

Tebriel
Tebriel

Reputation: 31

If I understand what your'e asking, this code doesn't do what you're thinking.

for department in store.departments ? []

It looks like you're using the existential operator ? similar to how you would a ternary operator a?b:c. From coffeescript.org:

It's a little difficult to check for the existence of a variable in JavaScript. if (variable) ... comes close, but fails for zero, the empty string, and false. CoffeeScript's existential operator ? returns true unless a variable is null or undefined, which makes it analogous to Ruby's nil?

If I wanted to later use the names, I'd write something like:

if store.departments?
    names = (department.name for department in store.departments)

You could put it all on one line, but with a list comprehension, that becomes pretty unreadable. The existential operator will test null && undef and only return true if it really exists.

If you want to use a ternary operator in coffeescript, it's less terse:

for department in if store.departments? then store.departments else []

Maybe not exactly what you want, because it's extraordinarily verbose here.

Upvotes: 1

Kyle
Kyle

Reputation: 22258

What about this?

if store.departments  
  alert department.name for department in store.departments

Or

alert department.name for department in store.departments if store.departments

Both statements compile to:

var department, _i, _len, _ref;

if (store.departments) {
  _ref = store.departments;
  for (_i = 0, _len = _ref.length; _i < _len; _i++) {
    department = _ref[_i];
    alert(department.name);
  }
}

Upvotes: 2

Related Questions