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?

• RobG: 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.

Answer 1

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).

• RobG: Not sure that explains why one should be chosen over the other.
• bvgheluwe: Thanks, this was some (important) detail that I overlooked. This also explains why return new (ctor.bind(null, args)); would not suit.
• SLaks: @RobG: They do two different things. You should chose whichever one you need.
• RobG: The OP started with "I understand how Function.prototype.bind.apply works..." so I naively assumed that s/he knew how it works.
• bvgheluwe: Maybe I should rephrase it as "I think I understand how ... works".