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JavaScript apply()

When I pass an array and Object respectively to function.apply(), I get the o/p of NaN but when I do it Object and array, I get a Number. Why is this happening?
since arrays are also considered as objects, why I'm unable to use this to represent array from the same context/
I've tried changing the position of variable calling but I know for a fact that the order only matters when it's a parameter.
function add() {
return arr[0] + arr[1] + this.a + this.b;
}
let obj = {
a: 1,
b: 2
};
let arr = [1, 2];
console.log(add.apply(arr, obj));
console.log(add.apply(obj, arr));
O/P
NaN
6

Perhaps a diagram could help. Your function uses no parameters, so what you supply in the second argument to apply (which should be an array of values) doesn't matter. Only the first parameter (which becomes this in the function body) matters.
When you call with arr, then this points to the arr supplied in the argument, but arr also points to the same object, as it's global and not overwritten anywhere:
function add() {
// 1 + 1 + undefined + undefined //=> NaN
return arr[0] + arr[1] + this.a + this.b;
} // | | | |
// +------+--------+---------+-----------+
// |
// V
let arr = [1, 2]; let obj = {a: 1, b: 2};
// ^
// `-----Nothing points to this
add.apply(arr) // equivalent to `add.apply(arr, obj)`, since `add` ignores parameters
When you call with obj, then this points to the obj supplied in the argument, and arr again points to the global arr object:
function add() {
// 1 + 2 + 1 + 2 //=> 6
return arr[0] + arr[1] + this.a + this.b;
} // | | | |
// +------+--------+ +-----------+
// | |
// V V
let arr = [1, 2]; let obj = {a: 1, b: 2};
add.apply(obj) // equivalent to `add.apply(obj, arr)`, since `add` ignores parameters

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/apply
function.apply(thisArg, [argsArray]) your first argument becomes this in the function, and second argument is an array of arguments youre passing to the function
so add.apply(arr, obj) really should be add.apply(arr, [obj]) would translate to the following, which this(arr) doesnt have properties a nor b
function add() {
return arr[0] + arr[1] + arr.a + arr.b;
}
// then invoking it as
add(obj);
and add.apply(obj, arr) would translate to
function add() {
return arr[0] + arr[1] + obj.a + obj.b;
}
// them invoking it as
add(1, 2);

JavaScript compose functions

I am reading a book which contains the following example:
var composition1 = function(f, g) {
return function(x) {
return f(g(x));
}
};
Then the author writes: "...naive implementation of composition, because it does not take the execution context into account..."
So the preferred function is that one:
var composition2 = function(f, g) {
return function() {
return f.call(this, g.apply(this, arguments));
}
};
Followed by an entire example:
var composition2 = function composition2(f, g) {
return function() {
return f.call(this, g.apply(this, arguments));
}
};
var addFour = function addFour(x) {
return x + 4;
};
var timesSeven = function timesSeven(x) {
return x * 7;
};
var addFourtimesSeven2 = composition2(timesSeven, addFour);
var result2 = addFourtimesSeven2(2);
console.log(result2);
Could someone please explain to me why the composition2 function is the preferred one (maybe with an example)?
EDIT:
In the meantime i have tried to use methods as arguments as suggested, but it did not work. The result was NaN:
var composition1 = function composition1(f, g) {
return function(x) {
return f(g(x));
};
};
var composition2 = function composition2(f, g) {
return function() {
return f.call(this, g.apply(this, arguments));
}
};
var addFour = {
myMethod: function addFour(x) {
return x + this.number;
},
number: 4
};
var timesSeven = {
myMethod: function timesSeven(x) {
return x * this.number;
},
number: 7
};
var addFourtimesSeven1 = composition1(timesSeven.myMethod, addFour.myMethod);
var result1 = addFourtimesSeven1(2);
console.log(result1);
var addFourtimesSeven2 = composition2(timesSeven.myMethod, addFour.myMethod);
var result2 = addFourtimesSeven2(2);
console.log(result2);

This just answers what composition2 actually does:
composition2 is used when you want to keep this as context in the functions itself. The following example shows that the result is 60 by using data.a and data.b:
'use strict';
var multiply = function(value) {
return value * this.a;
}
var add = function(value) {
return value + this.b;
}
var data = {
a: 10,
b: 4,
func: composition2(multiply, add)
};
var result = data.func(2);
// uses 'data' as 'this' inside the 'add' and 'multiply' functions
// (2 + 4) * 10 = 60
But yet, it still breaks the following example (unfortunately):
'use strict';
function Foo() {
this.a = 10;
this.b = 4;
}
Foo.prototype.multiply = function(value) {
return value * this.a;
};
Foo.prototype.add = function(value) {
return value + this.b;
};
var foo = new Foo();
var func = composition2(foo.multiply, foo.add);
var result = func(2); // Uncaught TypeError: Cannot read property 'b' of undefined
Because the context of composition2 (this) is undefined (and is not called in any other way, such as .apply, .call or obj.func()), you'd end up with this being undefined in the functions as well.
On the other hand, we can give it another context by using the following code:
'use strict';
var foo = new Foo();
var data = {
a: 20,
b: 8,
func: composition2(foo.multiply, foo.add)
}
var result = data.func(2);
// uses 'data' as 'this'
// (2 + 8) * 10 = 200 :)
Or by explicitly setting the context:
'use strict';
var multiply = function(value) {
return value * this.a;
};
var add = function(value) {
return value + this.b;
};
var a = 20;
var b = 8;
var func = composition2(multiply, add);
// All the same
var result1 = this.func(2);
var result2 = func.call(this, 2);
var result3 = func.apply(this, [2]);

composition1 would not pass arguments other than the first to g()
If you do:
var composition1 = function(f, g) {
return function(x1, x2, x3) {
return f(g(x1, x2, x3));
}
};
the function will work for the first three arguments. If you however want it to work for an arbitrary number, you need to use Function.prototype.apply.
f.call(...) is used to set this as shown in Caramiriel's answer.

I disagree with the author.
Think of the use-case for function-composition. Most of the time I utilize function-composition for transformer-functions (pure functions; argument(s) in, result out and this is irrelevant).
2nd. Utilizing arguments the way he does it leads into a bad practice/dead end, because it implies that the function g() might depend on multiple arguments.
That means, that the composition I create is not composable anymore, because it might not get all arguments it needs.
composition that prevents composition; fail
(And as a side-effect: passing the arguments-object to any other function is a performance no-go, because the JS-engine can't optimize this anymore)
Take a look at the topic of partial application, usually misreferenced as currying in JS, wich is basically: unless all arguments are passed, the function returns another function that takes the remaining args; until I have all my arguments I need to process them.
Then you should rethink the way you implement argument-order, because this works best when you define them as configs-first, data-last.Example:
//a transformer: value in, lowercased string out
var toLowerCase = function(str){
return String(str).toLowerCase();
}
//the original function expects 3 arguments,
//two configs and the data to process.
var replace = curry(function(needle, heystack, str){
return String(str).replace(needle, heystack);
});
//now I pass a partially applied function to map() that only
//needs the data to process; this is really composable
arr.map( replace(/\s[A-Z]/g, toLowerCase) );
//or I create another utility by only applying the first argument
var replaceWhitespaceWith = replace(/\s+/g);
//and pass the remaining configs later
arr.map( replaceWhitespaceWith("-") );
A slightly different approach is to create functions that are, by design, not intended to get all arguments passed in one step, but one by one (or in meaningful groups)
var prepend = a => b => String(a) + String(b); //one by one
var substr = (from, to) => value => String(str).substr(from, to); //or grouped
arr.map( compose( prepend("foo"), substr(0, 5) ) );
arr.map( compose( prepend("bar"), substr(5) ) );
//and the `to`-argument is undefined; by intent
I don't intend to ever call such functions with all the arguments, all I want to pass them is their configs, and to get a function that does the job on the passed data/value.
Instead of substr(0, 5, someString), I would always write someString.substr(0, 5), so why take any efforts to make the last argument (data) applyable in the first call?

How to turn an undefined var to an object, inside a function

There is something I can't find an answer or an explanation for. Let's take for example the following code:
function fn(x){
x = {value: 10};
}
var a;
fn(a);
alert(a.value); //a is undefined
Shouldn't a = {value: 10}; as we passed it through that function?

The x is locally scoped. You are passing only values and not references. So you might need to return and assign like this:
function fn(x){
x = {value: 10};
return x;
}
var a;
a = fn(a);
From an awesome article:
When passing in a primitive type variable like a string or a number, the value is passed in by value. This means that any changes to that variable while in the function are completely separate from anything that happens outside the function.
function myfunction(x)
{
// x is equal to 4
x = 5;
// x is now equal to 5
}
var x = 4;
alert(x); // x is equal to 4
myfunction(x);
alert(x); // x is still equal to 4
Passing in an object, however, passes it in by reference. In this case, any property of that object is accessible within the function.
function myobject()
{
this.value = 5;
}
var o = new myobject();
alert(o.value); // o.value = 5
function objectchanger(fnc)
{
fnc.value = 6;
}
objectchanger(o);
alert(o.value); // o.value is now equal to 6

double double parentheses javascript

I found this code in a book:
function foo() {
console.log( this.a );
}
var a = 2;
var o = { a: 3, foo: foo };
var p = { a: 4 };
o.foo(); // 3
(p.foo = o.foo)(); // 2
What does last line mean?

The last line is doing an assignment and then calling the function.
Assignment happens first
(p.foo = o.foo)
Then call the function
(p.foo = o.foo)();
In this second call to foo, it is being called outside of the scope of p or o, so it's essentially the same as calling:
foo();

How to change a function accepted arguments after creating it on Javascript?

Or, in other words, how to make this work:
function foo(){}
//do something that modifies foo as if it was defined with "function foo(a,b,c){};"
console.log(foo.length);
//output: 3

It is possible, but maybe not very nice:
function lengthDecorator(fun) {
function update(len) {
var args = []; // array of parameter names
for (var i = 0; i < len; ++i) {
args.push('a' + i);
}
var result = new Function('fun',
'return function(' + args.join(',') + ') {' +
'var args = Array.prototype.slice.call(arguments);' +
'return fun.apply(this, args);' + // call supplied function
'}'
); // create a function that will return a function
result = result(fun); // make the fun param known to the inner function
result.update = update;
return result;
}
return update(fun.length);
}
Example usage:
var foo = lengthDecorator(function(a,b) {
return a+b;
});
print('foo.length: ' + foo.length);
print('foo(2, 3): ' + foo(2, 3));
print('');
foo = foo.update(42);
print('foo.length: ' + foo.length);
print('foo(2, 3): ' + foo(2, 3));
Output:
foo.length: 2
foo(2, 3): 5
foo.length: 42
foo(2, 3): 5
(Live demo: Ideone.com, jsFiddle)
lengthDecorator wraps the supplied function with a function that takes the same amount of parameters as the supplied function. The parameter count can be changed with update.
C.f.
new Function(...): Dynamically create a new function.
fun.apply(...): "Calls a function with a given this value and arguments provided as an array."

function foo() {}
alert(foo.length); // 0
foo = function (a, b, c) {}
alert(foo.length); // 3

I'm not sure what you're actually trying to do, but you can store the old foo in var and then redefine foo.
function foo() {...}
var oldfoo = foo;
foo = function (a, b, c) {
oldfoo();
}
But what's the point?

The length property of a function object is non-writable and non-configurable, so there is no way to change its value.
You could define a new function which invokes the original function internally...