Why Function.prototype.bind.apply?

Question

I understand how Function.prototype.bind.apply works. Suppose the following definitions:

var Constructor = function(someArgument){
    //constructor body here
};

var creator = function(ctor, args){
    args.unshift(null); //or anything else to make 'bind' succeed
    return new (Function.prototype.bind.apply(ctor, args));
};

var creator2 = function(ctor, args){
    return new ctor(args);
};

var obj = creator(Constructor, ['someVar']);
var obj2 = creator2(Constructor, ['someVar']);

Both creator and creator2 work, so why would one choose creator over creator2?

apply is just a convenient way to set the value of this when calling function so you can just pass arguments (for example) rather than having to build up a call expression containing each argument. I don't think there is anything to recommend it other than convenience. If there are zero or one parameters, then it doesn't have any benefit over call. — RobG
– RobG, Commented Jan 18, 2016 at 23:00

SLaks · Accepted Answer · 2016-01-18 22:54:41Z

2

Calling creator2 passes the args array as a single array parameter to the constructor.

Calling creator passes each item in the array as a separate parameter (via apply).

answered Jan 18, 2016 at 22:54

SLaks

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5 Comments

RobG Over a year ago

Not sure that explains why one should be chosen over the other.

bvgheluwe Over a year ago

Thanks, this was some (important) detail that I overlooked. This also explains why return new (ctor.bind(null, args)); would not suit.

SLaks Over a year ago

@RobG: They do two different things. You should chose whichever one you need.

RobG Over a year ago

The OP started with "I understand how Function.prototype.bind.apply works…" so I naively assumed that s/he knew how it works.

bvgheluwe Over a year ago

Maybe I should rephrase it as "I think I understand how ... works".

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