puzzling javascript array syntax

Question

I'm puzzled by this construction. It's from a published npm module. It seems the author knows his javascript. Versions of it appear several times in the module.

[].concat( opts['boolean'] ).filter( Boolean ).forEach( function( key ){
  flags.bools[key] = true;
});

Since it doesn't modify the original array and also the member 'boolean' is known from the outset, why not just:

opts.boolean.filter(Boolean).forEach(....

opts['boolean'] probably isn't an Array, so you can't filter or map over it. — Hunter McMillen
– Hunter McMillen, Commented Mar 28, 2017 at 19:45
While the author probably wants a shorthand to avoid calling .concat on non-genuine Array objects, the .filter() will keep any truthy values, so if opts.boolean is a plain object, it will be doing flags.bools["[object Object]"] = true;. Maybe the author is OK with that, but it doesn't seem like the best approach. — user1106925
– user1106925, Commented Mar 28, 2017 at 20:01
IMO, a better way would be if (Array.isArray(opts.boolean)) { opts.boolean.filter(Boolean).forEach(...) } — user1106925
– user1106925, Commented Mar 28, 2017 at 20:04
I guess I should have mentioned that opts.boolean is an optional user supplied value. When present it must be an Array but it may be undefined. — AdamW
– AdamW, Commented Mar 28, 2017 at 20:51
I had to look this one up. Boolean called w/o new returns a primitive true/false value depending on the truthiness of the argument. so the .filter returns all truthy entries in opts.boolean — AdamW
– AdamW, Commented Mar 28, 2017 at 20:58

AlexeyKuznetsov · Accepted Answer · 2017-03-28 19:50:12Z

0

Maybe because of this:

[].concat(undefined) // => [undefined]

undefined.filter(...) // => ERROR

answered Mar 28, 2017 at 19:50

AlexeyKuznetsov

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