Purpose of call with null in curry function (JavaScript implementation) - javascript

In the following implementation of curry function here:
// curry :: ((a, b, ...) -> c) -> a -> b -> ... -> c
function curry(fn) {
const arity = fn.length;
return function $curry(...args) {
if (args.length < arity) {
return $curry.bind(null, ...args);
}
return fn.call(null, ...args);
};
}
what's the purpose of using call with null as this in the fn.call? Wouldn't just fn(...args) work? I did try some examples and got the same results.

Yes, it would work just fine. (The tiny difference is that it passes undefined instead of null for the this context).
So why did the author use .call(null, …)? Probably because it nicely mirrors the .bind(null, …) call.

Related

What does "a.call.apply(a.bind, arguments)" mean?

Just because of curiosity I wanted to make a bounded function with this particular approach :
var fn = function(a, b, c) {
return a.call.apply(a.bind, arguments)
}
var boundFn = function(a, b, c) {
fn.apply(null, arguments)
}
function unboundedFn() {
console.log(this, arguments)
}
var boundedFn = boundFn(unboundedFn, x, y);
So I'm trying to understand what a.call.apply(a.bind, arguments) do exactly ?
If You have a function like this:
function unboundedFn() {
console.log(this, arguments)
}
You can use unboundedFn.call(thisObj, arg1, arg2) or unboundedFn.apply(thisObj, [arg1, arg2]) to run it, but changing what this means inside. Both call and apply do the same, and only difference is way of passing arguments.
Becouse call, apply and also bind are methods, you can run for example unboundedFn.call.call.call.call.apply.apply.apply.apply(), but it doesn't seem to have to much sense.
In your example return a.call.apply(a.bind, arguments) is equal to return a.bind.call(...arguments), which is equal to a.bind(...arguments.slice(1)), so whole fn function can be simplified to:
function fn(a,b,...args){
return a.bind(b, ...args);
}

How to pass a function with arguments to a function as argument in javascript? [duplicate]

This question already has answers here:
Passing a function with parameters as a parameter?
(7 answers)
Closed 3 years ago.
I'm new to javascript and I was learning higher-order functions where I learned about passing functions as arguments to other functions. How can I pass a function with arguments to a function?
I want to pass the function's arguments when passing the function as parameter to the main function.
example :
function fubar(a,b,fun(c,d)){
//execute here
}
and not like this
function fubar(a,b,fun()){
fun(a,b)
}
Here's an example, just pass it like a regular argument, no specific syntax is required :)
// define the function argument like a regular argument
function fubar(a, b, fn){
// call that argument like a function
return fn(a, b);
}
// sample function
function add(c, d){
return c + d;
}
// pass the "add" function like a regular argument
let result = fubar(1, 2, add);
console.log(result);
You just need to pass the name of the function as the parameter, just like another variable.
for example :
foobar(5, 10, baz, ham);
function foobar(a, b, fn1, fn2){
...
fn1(); // call the baz function.
res = fn2(a,b); // calls ham func with a and b as parameter
console.log(res)
...
}
function baz(){
console.log("Inside the Baz function ");
}
function ham(s,t){
console.log(s, t);
return s+t;
}
If you're asking how to pass a function that will execute with some pre-determined arguments, you can do so using another function.
For example
const a = 'a'
const b = 'b'
const c = 'c'
const d = 'd'
function fubar(a, b, fn) {
console.info('func:', a, b, fn()) // executing "fn()" here
}
function fun(arg1, arg2) {
console.info('fun:', arg1, arg2)
return 'return value from fun' // just an example so you can see it in fubar()
}
fubar(a, b, () => fun(c, d))
Here, () => fun(c, d) is an self-contained, anonymous arrow function (or "lambda") that when called, will execute fun(c, d).

Why is it necessary to pass existing arguments to `setTimeout` again?

Why is it necessary to write a and b after mls in the setTimeout method here? Aren’t a and b already defined in the arrow function?
function f(a, b) {
alert(a + b);
}
// shows 3 after 1 second
Function.prototype.defer = function(mls) {
return (a, b) => setTimeout(this, mls, a, b);
}
f.defer(1000)(1, 2);
Because f.defer(1000) returns the arrow function which in turn gets called with (1,2) as parameter. This arrow function is calling setTimeout which is supposed to call the same function f with these parameters 1 and 2. And the definition of seTimeout states that you can pass parameter to the callback function after first 2 parameters. Please refer below definition and link for the reference.
window.setTimeout(function[, delay, param1, param2, ...]);
https://developer.mozilla.org/en-US/docs/Web/API/WindowOrWorkerGlobalScope/setTimeout
The arrow function is defining a function that then invokes setTimeout(this, mls, a, b). It could also be written as:
Function.prototype.defer = function(mls){
var that = this;
return function(a,b) {
setTimeout(that, mls, a, b);
}
}
This is important to note because you're actually returning a function from defer. So when you call f.defer(1000), the return value is a function - your arrow function, to be specific - and then you invoke that function with the values 1 and 2 when you call f.defer(1000)(1, 2);.
If you don't, your function will not be called with any parameters. These parameters are passed into the originally invoked function.
setTimeout takes a variable number of arguments. The first one is the function (this, or f in this case), the second one is the delay. Any argument after that is passed to the to-be-called function upon calling it.
Your function f expects two arguments; if they aren’t passed, they are undefined. There are indeed an “a and b already defined in the arrow function”, but they are scoped to the arrow function. If a function calls another function, its arguments aren’t automatically transfered.
Internally, setTimeout is doing something like outerWorking in this example:
const inner = (a, b) => a + b,
outerWorking = (a, b) => inner(a, b),
outerNotWorking = (a, b) => inner();
console.log(outerWorking(2, 3)); // 5
console.log(outerNotWorking(2, 3)); // NaN
So f.defer(1000) returns a new function with mls === 1000 and this === f. It accepts a and b which are yet to be defined with the next function call (calling the arrow functions, with the argument list (1, 2)). Only when you pass a, b at f.defer(1000)(1, 2), then setTimeout(this, mls,a,b) will be called with a === 1, b === 2 and the other two bindings above (this and delay).
Alternatively, this could be written as
Function.prototype.defer = function(mls, a, b){
setTimeout(this, mls, a, b);
};
f.defer(1000, 1, 2)
But there’s an advantage in the original, curried approach, as you can simply detach the “defered function” from the final call:
const add1sDelay = f.defer(1000);
add1sDelay(1, 2);

Can someone explain this "passing arguments" in javascript example?

I'm reading through Javascript Garden, and I'm trying to wrap my head around the following example:
Passing Arguments
The following is the recommended way of passing arguments from one function to another.
function foo() {
bar.apply(null, arguments);
}
function bar(a, b, c) {
// do stuff here
}
Another trick is to use both call and apply together to create fast, unbound wrappers.
function Foo() {}
Foo.prototype.method = function(a, b, c) {
console.log(this, a, b, c);
};
// Create an unbound version of "method"
// It takes the parameters: this, arg1, arg2...argN
Foo.method = function() {
// Result: Foo.prototype.method.call(this, arg1, arg2... argN)
Function.call.apply(Foo.prototype.method, arguments);
};
I'm trying to figure out two things:
1) What exactly is an "unbound wrapper"?
2) How does the chaining from .call to .apply work and/or make the code faster?
"1) What exactly is an "unbound wrapper"?"
Unbound wrapper is just a function that calls another function by passing on the desired this value and arguments.
"2) How does the chaining from .call to .apply work and/or make the code faster?"
The .call.apply() is faster than having to do a .slice() on the arguments to separate the this from the actual args.
Otherwise it would need to do this, which is slower:
Foo.method = function(ths) {
Foo.prototype.method.apply(ths, Array.prototype.slice.call(arguments, 1));
};
What exactly is an "unbound wrapper"?
A function that is not to be called on an instance, but with an instance as its argument. It's not bound to the prototype / doesn't need to be bound to an instance. Example:
var x = new Foo;
// instead of
x.method(1, 2, 3);
// you now call
Foo.method(x, 1, 2, 3);
The benefit of this is that you can pass the function around without caring about its this context.
How does the chaining from .call to .apply work and/or make the code faster?
It doesn't really make anything "faster". It's not even compared to any "slower" solution.
For how it works, please check the duplicate question What's the meaning to chain call and apply together?.
I think the code should be like this:
function Foo() {}
Foo.prototype.method = function(a, b, c) {
console.log(this, a, b, c);
};
Foo.method = function() {
//Notice this line:
Function.apply.call(Foo.prototype.method, this, arguments);
};
then
Foo.method(1,2,3) => function Foo() {} 1 2 3
Other examples:
Function.apply.call(Array,this,[1,2]) => [1, 2]
Function.call.apply(Array,this,[1,2]) => [window]
Function.call.call(Array,this,[1,2]) => [[1, 2]]

I'm reading Eloquent Javascript and I am a little confused by this partial function example. Please help explain

function asArray(quasiArray, start) {
var result = [];
for (var i = (start || 0); i < quasiArray.length; i++)
result.push(quasiArray[i]);
return result;
}
function partial(func) {
var fixedArgs = asArray(arguments, 1);
return function(){
return func.apply(null, fixedArgs.concat(asArray(arguments)));
};
}
function compose(func1, func2) {
return function() {
return func1(func2.apply(null, arguments));
};
}
var isUndefined = partial(op["==="], undefined);
var isDefined = compose(op["!"], isUndefined);
show(isDefined(Math.PI));
show(isDefined(Math.PIE));
Why can't the function compose simply return:
func1(func2);
and give the proper output. I thought the partial function which is stored in the variable isUndefined already returns func.apply(null, [fixed, arguments])
var op = {
"+": function(a, b){return a + b;},
"==": function(a, b){return a == b;},
"===": function(a, b){return a === b;},
"!": function(a){return !a;}
/* and so on */
};
Both partial and compose are higher-order functions.
isUndefined will return a function that, when invoked, will invoke the originally passed function with the original arguments plus any new arguments passed at invocation.
To answer your question, you'd be calling apply on the function returned from partial which will in turn, call apply on the function originally passed to partial.
You want compose to return a function that when called, will return the result of calling the first function passed the second function as an argument (with the second function passed the arguments passed to the compose invocation). If compose returned func1(func2), then you'd assign the result of the invocation to the variable isDefined.
EDIT:
Now that we have op, let's try to decompose this:
var isUndefined = partial(op["==="], undefined);
this is equivalent to
var isUndefined = partial(function(a, b){return a === b;}, undefined);
isUndefined is assigned a function that, when called, will call the function passed as the first argument to partial, passing in undefined as the first argument to that function call, followed by the arguments passed to the function isUndefined i.e.
partial(function(a, b){return a === b;}, undefined /* this will become 'a' when isUndefined is invoked */)(argumentForisUndefined /* this will become 'b' when isUndefined is invoked */);
isDefined composes isUndefined with another function that negates the result of isUndefined.
var isDefined = compose(op["!"], isUndefined);
is equivalent to
var isDefined = compose(function(a){return !a;}, isUndefined);
which is equivalent to (renamed variables for clarity)
var isDefined = compose(
function(a){return !a;},
partial( /* partial function becomes 'a' passed to first function */
function(b, c) {
return b === c;
},
undefined /* undefined becomes 'b' passed to partial */
)
)(argumentForisDefined /* argumentForisDefined becomes 'c' passed to partial */);
If we look at what we have so far and substituting for readability, boils down to a function that takes an argument and compares it to undefined, negates the result and returns a boolean
var isDefined = function (b) { return !undefined === b; }
So lets simply dissect it. Assuming we have this compose function:
function compose(func1, func2) {
return func1(func2.apply(null, arguments));
}
What will happen when you use it like this?
a = compose(function(){console.log(1)}, function(){console.log(2)});
The second function would be call immediately outputting 2, and straight afterwards the first function will be called outputting 1. a will be undefined, because the first function does not return anything.
What you want combine to do, is to return a new function, that combines the two other functions and that you can call at will.
Doing the above all on the original compose, will return a new function, that, when you call it with a() will output 2 and then 1.

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