arrays - JavaScript: unsure how call and apply relate to the arguments object in this function -

in abstract way, totally understand what's going on. great way have unlimited number of arguments functions being passed calculate object.

var calculate = function(){      var fn = array.prototype.pop.apply(arguments);     return fn.apply(null, arguments);  },  sum = function(x,y){      return x + y;  },  diff = function(x,y){      return x - y;  }; 

this appears crux of function. using apply method here allow arguments object have pop method of array prototype

where unclear is...

the tut says going give function object , assign fn variable, remove function object arguments object, because of pop method. (the pop method removes last item of array, , assigns ever called pop method) in case fn variable. end arguments object, no longer has function object, has numeric values

var fn = array.prototype.pop.apply(arguments);         return fn.apply(null, arguments); 

recently, started realize perhaps misunderstanding seeing fn variable that, , not function passed calculate object.

thanks in advance helping me understand this!!

this syntax of apply:

function.apply(thisarg, [argsarray]) 

let's use calculate on sum:

calculate(10, 15, sum); 

inside calculate:

in start - arguments = [10, 15, sum]; note arguments not array, array object - using array notation easier

var fn = array.prototype.pop.apply(arguments);  

arguments not array, doesn't have pop method, take pop method array prototype , run on arguments - means use arguments 'this' prop of .pop(). returns last parameter of arguments, supposed function (sum in case).

after pop - arguments = [10, 15];

now take arguments , use 2nd part of .apply(thisarg, [argsarray]) method - argsarray.

in fn.apply(null, arguments) thisarg = null (doesn't matter function sum doesn't use this), , apply assigns arguments parameters in 'array' order - x (1st) = 10, y (2nd) = 15, , runs function.