JavaScript function composition by chaining - javascript

I have checked the possibility of duplicate question,
and cannot find the exact solution.
I wrote some function chain code in JavaScript as below, and works fine.
var log = function(args)
{
console.log(args)
return function(f)
{
return f;
};
};
(log('1'))(log('2'))(log('3'))(log('4'));
//1
//2
//3
//4
I want to make this lazy evaluation.
Or to compose function.
var log = function(args)
{
var f0 = function()
{
return console.log(args);
};
return function(f1)
{
return function()
{
f0();
return f1;
};
};
};
var world = (log('1'))(log('2'))(log('3'))(log('4'));
console.log(world);
//should be just a function,
// but in fact
//1
//[function]
world();
//should be
//1
//2
//3
//4
// but in fact
// 2
Something is very wrong.
Can you fix it?
Thanks.
This question is resolved, but there is further
async issue as shown in comment discussion
When we have
// unit :: a -> IO a
var unit = function(x)
{
return function()
{
return x;
};
};
// bind :: IO a -> (a -> IO b) -> IO b
var bind = function(x, y)
{
return function()
{
return y(x())();
};
};
// seq :: IO a -> IO b -> IO b
var seq = function(x, y)
{
return function()
{
return x(), y();
};
};
var action = function(x)
{
return function(y)
{
return y ? action(seq(x, y)) : x();
};
};
var wrap = function(f)
{
return function(x)
{
return action(function()
{
return f(x);
});
};
};
var log = wrap(console.log);
// -- runtime --
// HACK: when `world` is modified by passing a function,
// the function will be executed.
Object.defineProperties(window,
{
world:
{
set: function(w)
{
return w();
}
}
});
We also often want async chain reactions badly.
var asyncF = function(callback)
{
setTimeout(function()
{
for (var i = 0; i < 1000000000; i++)
{
};
callback("async process Done!");
}, 0);
};
var async = wrap(asyncF(function(msg)
{
world = log(msg);
return msg;
}));
Now,
world = (log(1))(async)(log(3));
//1
//3
//async process Done!
So far nice and smooth, now we try to use bind
world = (log(1))
(bind((async), (log(x))));
//should be
//1
//async process Done!
//3
//in fact
//ReferenceError: x is not defined
Could you modify to make this work, please?
one more about retrun x, y; multiple value
I don't understand
// seq :: IO a -> IO b -> IO b
var seq = function(x, y)
{
return function()
{
return x(), y();
};
};
as the library author mentions
Note that this is not possible in Haskell because one function can't return two results. Also, in my humble opinion, it looks ugly.
I agree, and don't know what this
return x(), y();
multiple return value.
I googled and searched here, but could not find an answer.
What is this??
(just in case, I will chose this hack for the syntax)
Thanks!

So if I understand the question correctly, you want to chain IO actions in JavaScript. To do so, you first need to define what an IO action is. One way to think of an IO action is that it is simply a function which takes no arguments. For example:
// log :: a -> IO b
function log(x) {
return function () { // IO action
return console.log(x);
};
}
One advantage of representing an IO action as a function with no arguments is that it is the same representation for thunks (unevaluated expressions). Thunks are the things that enable lazy evaluation in languages like Haskell. Hence you get laziness for free.
Now composition. How do you compose two IO actions in JavaScript? In Haskell, you use the >> operator to sequence IO actions, which is usually defined in terms of >>= (a.k.a. bind) as follows:
(>>=) :: Monad m => m a -> (a -> m b) -> m b
(>>) :: Monad m => m a -> m b -> m b
x >> y = x >>= \_ -> y
It is easy to write an equivalent bind function for our IO actions in JavaScript:
// bind :: IO a -> (a -> IO b) -> IO b
function bind(x, y) {
return function () {
return y(x())();
};
}
Suppose you have an IO action x :: IO a. Since it's just a function with no arguments, when you call it it's equivalent to evaluating the IO action. Hence x() :: a. Feeding this result to the function y :: a -> IO b results in the IO action y(x()) :: IO b. Note that the entire operation is wrapped in a superfluous function for laziness.
Similarly, it is just as straightforward to implement the >> operator. Let's call it seq as in “sequence”.
// seq :: IO a -> IO b -> IO b
function seq(x, y) {
return function () {
return x(), y();
};
}
Here we evaluate the IO expression x, don't care about its result and then return the IO expression y. This is exactly what the >> operator does in Haskell. Note that the entire operation is wrapped in a superfluous function for laziness.
Haskell also has a return function which lifts a value into a monadic context. Since return is a keyword in JavaScript, we'll call it unit instead:
// unit :: a -> IO a
function unit(x) {
return function () {
return x;
};
}
As it turns out there's also a sequence operator in Haskell which sequences monadic values in a list. It can be implemented in JavaScript for IO actions as follows:
// sequence :: [IO a] -> IO [a]
function sequence(array) {
return function () {
var list = array;
var length = list.length;
var result = new Array(length);
var index = 0;
while (index < length)
result[index] = list[index++]();
return result;
};
}
That's all we need. Now we can write:
var world = sequence([log("1"), log("2"), log("3"), log("4")]);
world();
// 1
// 2
// 3
// 4
Hope that helps.
Yes, it is indeed possible to chain IO actions using your syntax. However, we'll need to redefine what an IO action is:
function action(x) {
return function (y) {
return y ? action(seq(x, y)) : x();
};
}
Let's understand what the action function does using an example:
// log :: a -> IO b
// log :: a -> IO r -> IO r
function log(x) {
return action(function () {
return console.log(x);
});
}
Now you can do:
log("1")(); // :: b
log("1")(log("2")); // :: IO r
In the first case we evaluated the IO action log("1"). In the second case we sequenced the IO actions log("1") and log("2").
This allows you to do:
var world = (log("1"))(log("2"))(log("3"))(log("4"));
world();
// 1
// 2
// 3
// 4
In addition you can also do:
var newWorld = (world)(log("5"));
newWorld();
// 1
// 2
// 3
// 4
// 5
And so on....
Everything else remains the same. Note that this is not possible in Haskell because one function can't return two results. Also, in my humble opinion, it looks ugly. I prefer using sequence instead. However, this is what you wanted.

Let's look at what happens here:
var log = function(args)
{
var f0 = function()
{
return console.log(args);
};
return function(f1)
{
return function()
{
f0();
return f1;
};
};
};
And inline just a bit:
var log = function(args) {
return function(f1) {
return function() {
console.log(args);
return f1;
};
};
};
So we're returning a function f which accepts a function f1, and returns a function g which does logic and returns f1. Quite a mouthful! Your question is why does
(log('1'))(log('2'))(log('3'));
Log 1. I did away with the log('4') since going to 3 is enough to show your described case. To answer that let's play compiler and do the inlining game!
(log('1'))(log('2'))(log('3'))
// =>
(
function (f1) {
return function () {
console.log('1');
return f1;
}
}
)(
function (f1) {
return function () {
console.log('2');
return f1;
}
}
)(
function (f1) {
return function () {
console.log('3');
return f1;
}
}
)
Simple substitution. I took each instance of log(something), replaced it with the contents of the function, replaced the argument with the value passed. Let's do it again!
(
function () {
console.log('1');
return function (f1) {
return function () {
console.log('2');
return f1;
}
};
}
)(
function (f1) {
return function () {
console.log('3');
return f1;
}
}
)
This one is a bit trickier to follow: I expanded the first function call. The topmost function received an argument f1 to which we just supplied a value, so I went into the function and replaced every occurrence of f1 with the given value (the result of log('2')), just like with the log argument.
Look over what happened here again if you still don't follow, but my suggestion is doing it yourself: Copy the snippet into your favourite code editor and do the expansion yourself.
Now you may see why the log('1') was called. The next thing we, the compiler, need to do is take care of the next function call. And whadya know, the first line in that function is a console.log! Better do it!
What can we do!?
I don't know Haskell or the IO Monad, but as you currently planned, I do not think you can do what you want with basic functions, not like that. If you can say what problem you want to solve using this...erm...pattern, maybe we can help!

This is because you are just returning and returning everything...
There are three thing printed in the output:
1
function ()
{
f0();
return f1;
}
2
1) first output: 1
this is because:console.log(args) is only executed once in your chaining, because f0 is executed only once at the last where it finds args to be 1(because of returning each nested function, what value you return at the last is a function f1 which executes f0() when the value of args is 1.
then it prints 1 to the console.
2) second output the function f1
the return f1;(which is returned to the function when you passed args as 1) executed at the last returns
function ()
{
f0();
return f1;
}
back to the variable world, hence only the inner nested function is printed to the console.
3) Third output: 2
then when you execute the functionworld(),
Again the function f1 is executed directly(see there is just a little difference between world and world()) but this time the returned function for when you passed args as 2.
Reason: world will output only the function, world() will execute the function.
When you write world(), at the last time where the function f1 is returned the value for args was 2 is directly executed.
I know I have terribly worded the answer.. but hope this helps (hope you understand)

While executing
var world = (log('1'))(log('2'))(log('3'))(log('4'));
(log('1')) is executed first which returns a function which takes in (log('2')).
This anonymous function starts executing but doesn't accept any argument.
log('3') is neglected. This can verified by
if(typeof(arguments[0]) == 'function'){
console.log("Got a neglected argument");
console.log(arguments[0]());
}
After executing f0(); (which prints 1 to screen), we return f1 which points to function returned by log('2'), this takes in log('4');
This can be verified by doing:
world()()()
this outputs:
2
4
undefined

Related

How can I access array in function composition created by reduce? [duplicate]

I came across this pattern in redux compose function. I still don't understand how in the example below the functions are evaluated starting from the last and not from the first:
function f2(a) {
return a + a;
}
function f3(a) {
return a + a + a;
}
function f4(a) {
return a + a + a + a;
}
function f5(a) {
return a + a + a + a + a;
}
function compose(...funcs) {
return funcs.reduce(function x(a, b) {
return function y(...args) {
const temp = a(b(...args));
return temp;
};
});
}
const composedFunction = compose(f2, f3, f4, f5);
const result = composedFunction(2);
In the first reduce iteration the accumulator is f2 so we'll get f2(f3(2))=12. In the next iteration we'll call f4(12)=48. In the last iteration we'll call f5(48)=240. So the evaluation order is f5(f4(f2(f3(2)))). But using console.log I see that the evaluation order is f2(f3(f4(f5(2)))) which is also 240 by coincidence.
As far as I understand the function y is called for all array elements so why only the last function gets 2 as the parameter?
Let's step through the code with a very simple example:
compose(f2, f3, f4)
As no initial value was passed to reduce, it will start with the first (f2) and the second (f3) value of the array and call the callback with that, x gets called with a being f2 and b being f3. Now x does'nt do anything, it just returns function y that can access a and b through a closure.
Reduce will now continue to the third element, the first argument being the result of the previous callback (the closured y), and the second argument being f4. Now x gets called again, and another closure is created over y, y gets the finally returned from the whole function.
If we try to visualize thus closured function it'll be:
y { // closure of y
a -> y { // a references another closure of y
a -> f3,
b -> f2
},
b -> f4
}
Now you call that closured y and pass 2 into it, that will call b (f4) and pass the result to the call to a (closured y).
a ( b(...args))
y { ... } ( f4(2) )
Now that closured y will do the same:
a ( b ( ...args))
f2( f3( f4( 2 ) ) )
Hint: It is sometimes really difficult to keep track of closured values, therefore the console provides you with great utilities to keep track of them: Open your code in the consoles "debugger" tab, click on the line numbers where the function calls are to attach breakpoints, then run the code again, the execution will yield whenever a breakpoint is reached and you can see the values of all variables (including closured ones).
The reduce is not calling the functions f2, f3, f3, f5, but it is creating a function from those. This is the value of the accumulator in each iteration. Note that the value is a function and not a result from execution of the function.
1:a=f2;b=f3;return value(NOT TEMP but function y)=f2(f3(...args))
2:a(prev return value)=f2(f3(...args));b=f4;return value=f2(f3(f4(...args)))
and so on....
The compose function can be re-written as:
function compose(...funcs) {
return funcs.reduce(function (a, b) {
return function (arg) {
const temp = a(b(arg));
return temp;
};
});
}
After the first iteration, the returned function which is passed in as the next accumulator is:
function (arg) { // R1
return f2(f3(arg));
}
After the second iteration, the returned function which is passed in as the next accumulator is:
function (arg) { // R2
return R1(f4(arg));
}
And finally, the returned function assigned to composedFunction is:
function (arg) { // composedFunction
return R2(f5(arg));
}
So running composedFunction(2) and going back up the chain:
f5(2) returns 10
R2(10) returns R1(f4(10))
which is R1(40)
R1(40) returns f2(f3(40))
which is f2(120)
which is 240
Hopefully that's sufficient.
It can be written as a single call as:
function composedFunction(arg) {
return f2(f3(f4(f5(arg))));
}

NodeJS output is function instead of printed strings

I am having trouble printing the correct result in NodeJS, why isn't my code printing the strings in the correct way ? I feel like console.log is not called at all. Why do I get :
[Function]
[Function]
[Function]
[Function]
Expected result:
Tigrou (buddy of Spider) was here!
Spider (buddy of Tigrou) was also here!
Tigrou (buddy of Spider) are in a boat...
1 (buddy of 2)3
The code I thought would work:
function build_sentence(...args)
{
var i = 0
var s = ""
for (let arg of args)
{
if (i == 1)
{
i++
s += "(buddy of " + arg + ") "
}
else
{
s += arg + " "
i++
}
}
return s
}
function mycurry(build_sentence)
{
return function(...args)
{
if (!args)
{
return build_sentence();
}
else
{
return mycurry(build_sentence.bind(this, ...args))
}
}
}
const curried = mycurry(build_sentence);
console.log(curried("Tigrou")("Spider")(" was here!"))
console.log(curried("Spider")("Tigrou")(" was also here!"))
console.log(curried("Tigrou", "Spider", " are in a boat... "))
console.log(curried(1)(2, 3))
Here's a full working solution for you (except the string spacing)
const build_sentence_n_params = 3;
function build_sentence(...args)
{
var i = 0
var s = ""
for (let arg of args)
{
if (i == 1)
{
i++
s += "(buddy of " + arg + ") "
}
else
{
s += arg + " "
i++
}
}
return s
}
// A generic n-parameter curry helper, that returns fn return value
// after n-parameters given, and continues currying with more parameters.
// Note that in this configuration, the caller would also need
// to know how many parameters to give before a value is returned.
function ncurry(fn, n)
{
// Note that we have to return an outer function here, rather than just
// returning `curry` function directly, so `params` is unique for each
// initial call to `curried`.
return function(...args) {
// Initial arguments (note: we can use this array without copy, since
// rest params will always be a new array)
const params = args;
// This function contains the return logic without having to duplicate it below
function next() {
// If we have all params, call the given function, otherwise
// return the curry function again for more parameters
return !n || params.length >= n ? fn.call(null, ...params) : curry;
}
// The outer function is only called once, but this curry
// function will be called for each additional time.
function curry(...args) {
// Accumulate additional arguments
params.push(...args)
return next();
};
return next();
};
}
function mycurry(build_sentence)
{
// Call the generic n-parameter curry helper to generate
// a specific curry function for our purposes.
return ncurry(build_sentence, build_sentence_n_params);
}
const curried = mycurry(build_sentence);
console.log(curried("Tigrou")("Spider")(" was here!"))
console.log(curried("Spider")("Tigrou")(" was also here!"))
console.log(curried("Tigrou", "Spider", " are in a boat... "))
console.log(curried(1)(2, 3))
Your method invocations are quite confusing so I took the liberty to simplify your code.
If you run the following:
const curried = build_sentence("Tigrou", "Spider", " was here!");
console.log(curried);
You will get your desired output:
Tigrou (buddy of Spider) was here!
Why you are using the mycurry method is beyond my understanding.
When debugging your code, the sentence is already built on the first invocation, and what happens is that the subsequent invocations are not really invocations of the inner functions.
the main issue about our code is that !args is all the time false, as [] to boolean is also true.
you have to use !args.length. however you should add an extra call to your curried function usage like curried(1)(2, 3)().
the other approach is using comparison of curried function number of required params (build_sentence.length) and the number of params passed (args.length), but it's not working with spread scenario.
upd:
ex.1 - using Function.prototype.length property
const curry = fn =>
(...args) =>
fn.length <= args.length
? fn(...args)
: curry(fn.bind(this, ...args));
const func1 = (a, b, c) => `${a},${b},${c}`;
const func2 = (...args) => args.join(',');
console.log(curry(func1)(1,3)(4));
console.log(curry(func1)(1,3,4));
console.log(curry(func1)(1)(3)(4));
console.log(curry(func2)(1));
in this case currying will work fine for func1 (function that has enumerable number of arguments) because func1.length = 3. however for func2 - func2.length = 0, so (curry(func1)) will be executed after first call.
ex.2 - using number of arguments passed
const curry = fn =>
(...args) =>
!args.length
? fn()
: curry(fn.bind(this, ...args));
const func = (...args) => args.join(',');
console.log(curry(func)(1)(2,3,4)(5)());
in this case function currying will only return result of fn executing, when called with no arguments. however it will handle innumerable arguments of curried function properly.

JS Function With Two Parentheses and Two Params

I'm trying to understand how a function works that is run with two parentheses and two parameters. Like so:
add(10)(10); // returns 20
I know how to write one that takes two params like so:
function add(a, b) {
return a + b;
}
add(10,10); // returns 20
How could I alter that function so it could be run with one set of parameters, or two, and produce the same result?
Any help is appreciated. Literally scratching my head over this.
Thanks in advance!
How could I alter that function so it could be run with one set of parameters, or two, and produce the same result?
You can almost do that, but I'm struggling to think of a good reason to.
Here's how: You detect how many arguments your function has received and, if it's received only one, you return a function instead of a number — and have that function add in the second number if it gets called:
function add(a,b) {
if (arguments.length === 1) {
return function(b2) { // You could call this arg `b` as well if you like,
return a + b2; // it would shadow (hide, supercede) the one above
};
}
return a + b;
}
console.log(add(10, 10)); // 20
console.log(add(10)(10)); // 20
I said "almost" above because just because the add function received only one argument, that doesn't guarantee that the caller is going to call the result. They could write:
var x = add(10);
...and never call the function that x now refers to.
Welcome to the wonderful world of first order functions
In JavaScript, a function can return a function since a function is just another object. A simple implementation is something like:
function add(x){
return function addOther(y){
return x + y;
};
}
This is possible because of closures and first order functions.
This also lets you do partial application, libraries like Ramda utilize this to great extent.
var addThree = add(3)
addThree(5); // 8
To extend what both T. J. Crowder and Benjamin Gruenbaum said, libraries like Ramda (disclosure: I'm one of the authors) allow you to convert a simple function like this:
function add(a, b) {
return a + b;
}
into the style under discussion by wrapping it in a call to a curry function:
var add = R.curry(function add(a, b) {
return a + b;
});
add(3, 5); //=> 8
add(3)(5); //=> 8
var add3 = add(3);
add3(5); //=> 8
The best article I know on this subject is Hugh Jackson's Why Curry Helps. I wrote a more detailed one at Favoring Curry.
Update
Here is a version of curry somewhat simpler than the one in Ramda. It would do the above and quite a bit more, but doesn't do some of the things that Ramda does with placeholder values:
// here is a function that takes a function and returns a curried version
// of it, that is, a version that performs the sort of partial application
// you describe.
var curry = function(fn) {
// first, we detect how many arguments the function has.
var fnArity = fn.length;
var partialApply = function(args) {
// now, let's create a function that's curried
return function () {
// collect the previous args as the partial, and add the new
// ones you just received
var newArgs = (args || []).concat([].slice.call(arguments, 0));
// if we have "enough" arguments, we don't need any more partial
// application and we can call the function.
if (newArgs.length >= fnArity) {
return fn.apply(this, newArgs);
} else { // else we return a partially applied version
return partialApply(newArgs);
}
};
};
return partialApply([]); // a function is itself partially applied with 0 args
};
function add() {
var sum = 0;
for (var i = 0; i < arguments.length; i++) {
sum += arguments[i];
}
function total() {
for (var i = 0; i < arguments.length; i++) {
sum += arguments[i];
}
return total;
}
total.toString = function () { return sum };
return total;
}
This will work for any no of arguments and parentheses.
https://medium.com/#imdebasispanda/super-function-with-closure-86a58a9a980b

More succinct delayed evaluation than function(){return x}?

I'm porting some Python code that relies heavily on delayed evaluation. This is accomplished by via thunks. More specifically, any Python expression <expr> for which delayed evaluation is desired gets enclosed within a Python "lambda expression", i.e. lambda:<expr>.
AFAIK, the closest JavaScript equivalent of this is function(){return <expr>}.
Since the code I'm working with is absolutely awash in such thunks, I'd like to make the code for them more succinct, if at all possible. The reason for this is not only to save characters (a non-negligible consideration when it comes to JS), but also to make the code more readable. To see what I mean, compare this standard JavaScript form:
function(){return fetchx()}
with
\fetchx()
In the first form, the substantive information, namely the expression fetchx(), is typographically obscured by the surrounding function(){return...}. In the second form1, just one (\) character is used as "delayed evaluation marker". I think this is the optimal approach2.
AFAICT, solutions to this problem would fall into the following categories:
Using eval to simulate delayed evaluation.
Some special JavaScript syntax that I don't know about, and that accomplishes what I want. (My vast ignorance of JavaScript makes this possibility look quite real to me.)
Writing the code in some non-standard JavaScript that gets programmatically processed into correct JavaScript. (Of course, this approach will not reduce the final code's footprint, but may at least retain some gains in readability.)
None of the above.
I'm particularly interested in hearing responses of the last three categories.
P.S.: I'm aware that the use of eval (option 1 above) is widely deprecated in the JS world, but, FWIW, below I give a toy illustration of this option.
The idea is to define a private wrapper class whose sole purpose would be to tag plain strings as JavaScript code for delayed evaluation. A factory method with a short name (e.g. C, for "CODE") is then used to reduce, e.g.,
function(){return fetchx()}
to
C('fetchx()')
First, definitions of the factory C and of the helper function maybe_eval:
var C = (function () {
function _delayed_eval(code) { this.code = code; }
_delayed_eval.prototype.val = function () { return eval(this.code) };
return function (code) { return new _delayed_eval(code) };
})();
var maybe_eval = (function () {
var _delayed_eval = C("").constructor;
return function (x) {
return x instanceof _delayed_eval ? x.val() : x;
}
})();
The following comparison between a get function and a lazyget function shows how the above would be used.
Both functions take three arguments: an object obj, a key key, and a default value, and they both should return obj[key] if key is present in obj, and otherwise, the default value.
The only difference between the two functions is that the default value for lazyget can be a thunk, and if so, it will get evaluated only if key is not in obj.
function get(obj, key, dflt) {
return obj.hasOwnProperty(key) ? obj[key] : dflt;
}
function lazyget(obj, key, lazydflt) {
return obj.hasOwnProperty(key) ? obj[key] : maybe_eval(lazydflt);
}
Too see these two functions in action, define:
function slow_foo() {
++slow_foo.times_called;
return "sorry for the wait!";
}
slow_foo.times_called = 0;
var someobj = {x: "quick!"};
Then, after evaluating the above, and using (e.g.) Firefox + Firebug, the following
console.log(slow_foo.times_called) // 0
console.log(get(someobj, "x", slow_foo())); // quick!
console.log(slow_foo.times_called) // 1
console.log(lazyget(someobj, "x",
C("slow_foo().toUpperCase()"))); // quick!
console.log(slow_foo.times_called) // 1
console.log(lazyget(someobj, "y",
C("slow_foo().toUpperCase()"))); // SORRY FOR THE WAIT!
console.log(slow_foo.times_called) // 2
console.log(lazyget(someobj, "y",
"slow_foo().toUpperCase()")); // slow_foo().toUpperCase()
console.log(slow_foo.times_called) // 2
prints out
0
quick!
1
quick!
1
SORRY FOR THE WAIT!
2
slow_foo().toUpperCase()
2
1...which may strike Haskell programmers as strangely familiar. :)
2There's another approach, the one used, e.g., by Mathematica, that avoids the need for delayed evaluation markers altogether. In this approach, as part of a function's definition, one can designate any one of its formal arguments for non-standard evaluation. Typographically, this approach is certainly maximally unobtrusive, but a bit too much so for my taste. Besides, it is not as flexible, IMHO, as using, e.g., \ as a delayed evaluation marker.
In my humble opinion I think you're looking at this problem from a wrong perspective. If you're creating thunks manually then you need to consider refactoring your code. In most cases thunks should be:
Either returned from lazy functions.
Or created by composing functions.
Returning Thunks from Lazy Functions
When I first started practicing functional programming in JavaScript I was mystified by the Y combinator. From what I had read online the Y combinator was a divine entity to be worshipped. It somehow allowed functions which didn't know their own name to call themselves. Hence it was the mathematical manifestation of recursion - one of the most important pillars of functional programming.
However understanding the Y combinator was no easy feat. Mike Vanier wrote that the knowledge of the Y combinator is a diving line between those people who are "functionally literate" and those who aren't. Honestly, the Y combinator in itself is dead simple to understand. However most articles online explain it backwards making it difficult to understand. For example Wikipedia defines the Y combinator as:
Y = λf.(λx.f (x x)) (λx.f (x x))
In JavaScript this would translate to:
function Y(f) {
return (function (x) {
return f(x(x));
}(function (x) {
return f(x(x));
}));
}
This definition of the Y combinator is unintuitive and it doesn't make apparent how the Y combinator is a manifestation of recursion. Not to mention that it cannot be used at all in eager languages like JavaScript because the expression x(x) is evaluated immediately resulting in an infinite loop which eventually results in a stack overflow. Hence in eager languages like JavaScript we use the Z combinator instead:
Z = λf.(λx.f (λv.((x x) v))) (λx.f (λv.((x x) v)))
The resulting code in JavaScript is even more confusing and unintuitive:
function Z(f) {
return (function (x) {
return f(function (v) {
return x(x)(v);
});
}(function (x) {
return f(function (v) {
return x(x)(v);
});
}));
}
Trivially we can see that the only difference between the Y combinator and the Z combinator is that the lazy expression x(x) is replaced by the eager expression function (v) { return x(x)(v); }. It is wrapped in a thunk. In JavaScript however it makes more sense to write the thunk as follows:
function () {
return x(x).apply(this, arguments);
}
Of course here we're assuming that x(x) evaluates to a function. In the case of the Y combinator this is indeed true. However if the thunk doesn't evaluate to a function then we simply return the expression.
One of the most epiphanous moments for me as a programmer was that the Y combinator is itself recursive. For example in Haskell you define Y combinator as follows:
y f = f (y f)
Because Haskell is a lazy language the y f in f (y f) is only evaluated when required and hence you don't run into an infinite loop. Internally Haskell creates a thunk for every expression. In JavaScript however you need to create a thunk explicitly:
function y(f) {
return function () {
return f(y(f)).apply(this, arguments);
};
}
Of course defining the Y combinator recursively is cheating: you are just explicitly recursing inside the Y combinator instead. Mathematically the Y combinator itself should be defined non-recursively to describe the structure of recursion. Nonetheless we all love it anyway. The important thing is that the Y combinator in JavaScript now returns a thunk (i.e. we defined it using lazy semantics).
To consolidate our understanding let's create another lazy function in JavaScript. Let's implement the repeat function from Haskell in JavaScript. In Haskell the repeat function is defined as follows:
repeat :: a -> [a]
repeat x = x : repeat x
As you can see repeat has no edge cases and it calls itself recursively. If Haskell weren't so lazy it would recurse forever. If JavaScript were lazy then we could implement repeat as follows:
function repeat(x) {
return [x, repeat(x)];
}
Unfortunately if executed the above code would recurse forever until it results in a stack overflow. To solve this problem we return a thunk instead:
function repeat(x) {
return function () {
return [x, repeat(x)];
};
}
Of course since the thunk doesn't evaluate to a function we need another way to treat a thunk and a normal value identically. Hence we create a function to evaluate a thunk as follows:
function evaluate(thunk) {
return typeof thunk === "function" ? thunk() : thunk;
}
The evaluate function can now be used to implement functions which can take either lazy or strict data structures as arguments. For example we can implement the take function from Haskell using evaluate. In Haskell take is defined as follows:
take :: Int -> [a] -> [a]
take 0 _ = []
take _ [] = []
take n (x:xs) = x : take (n - 1) xs
In JavaScript we would implement take using evaluate as follows:
function take(n, list) {
if (n) {
var xxs = evaluate(list);
return xxs.length ? [xxs[0], take(n - 1, xxs[1])] : [];
} else return [];
}
Now you can use repeat and take together as follows:
take(3, repeat('x'));
See the demo for yourself:
alert(JSON.stringify(take(3, repeat('x'))));
function take(n, list) {
if (n) {
var xxs = evaluate(list);
return xxs.length ? [xxs[0], take(n - 1, xxs[1])] : [];
} else return [];
}
function evaluate(thunk) {
return typeof thunk === "function" ? thunk() : thunk;
}
function repeat(x) {
return function () {
return [x, repeat(x)];
};
}
Lazy evaluation at work.
In my humble opinion most thunks should be those returned by lazy functions. You should never have to create a thunk manually. However every time you create a lazy function you still need to create a thunk inside it manually. This problem can be solved by lifting lazy functions as follows:
function lazy(f) {
return function () {
var g = f, self = this, args = arguments;
return function () {
var data = g.apply(self, args);
return typeof data === "function" ?
data.apply(this, arguments) : data;
};
};
}
Using the lazy function you can now define the Y combinator and repeat as follows:
var y = lazy(function (f) {
return f(y(f));
});
var repeat = lazy(function (x) {
return [x, repeat(x)];
});
This makes functional programming in JavaScript almost as fun as functional programming in Haskell or OCaml. See the updated demo:
var repeat = lazy(function (x) {
return [x, repeat(x)];
});
alert(JSON.stringify(take(3, repeat('x'))));
function take(n, list) {
if (n) {
var xxs = evaluate(list);
return xxs.length ? [xxs[0], take(n - 1, xxs[1])] : [];
} else return [];
}
function evaluate(thunk) {
return typeof thunk === "function" ? thunk() : thunk;
}
function lazy(f) {
return function () {
var g = f, self = this, args = arguments;
return function () {
var data = g.apply(self, args);
return typeof data === "function" ?
data.apply(this, arguments) : data;
};
};
}
Creating Thunks by Composing Functions
Sometimes you need to pass expressions to functions that are evaluated lazily. In such situations you need to create custom thunks. Hence we can't make use of the lazy function. In such cases you can use function composition as a viable alternative to manually creating thunks. Function composition is defined as follows in Haskell:
(.) :: (b -> c) -> (a -> b) -> a -> c
f . g = \x -> f (g x)
In JavaScript this translates to:
function compose(f, g) {
return function (x) {
return f(g(x));
};
}
However it makes much more sense to write it as:
function compose(f, g) {
return function () {
return f(g.apply(this, arguments));
};
}
Function composition in mathematics reads from right-to-left. However evaluation in JavaScript is always from left-to-right. For example in the expression slow_foo().toUpperCase() the function slow_foo is executed first and then the method toUpperCase is called on its return value. Hence we want to compose functions in reverse order and chain them as follows:
Function.prototype.pipe = function (f) {
var g = this;
return function () {
return f(g.apply(this, arguments));
};
};
Using the pipe method we can now compose functions as follows:
var toUpperCase = "".toUpperCase;
slow_foo.pipe(toUpperCase);
The above code will be equivalent to the following thunk:
function () {
return toUpperCase(slow_foo.apply(this, arguments));
}
However there's a problem. The toUpperCase function is actually a method. Hence the value returned by slow_foo should set the this pointer of toUpperCase. In short we want to pipe the output of slow_foo into toUpperCase as follows:
function () {
return slow_foo.apply(this, arguments).toUpperCase();
}
The solution is actually very simple and we don't need to modify our pipe method at all:
var bind = Function.bind;
var call = Function.call;
var bindable = bind.bind(bind); // bindable(f) === f.bind
var callable = bindable(call); // callable(f) === f.call
Using the callable method we can now refactor our code as follows:
var toUpperCase = "".toUpperCase;
slow_foo.pipe(callable(toUpperCase));
Since callable(toUpperCase) is equivalent to toUpperCase.call our thunk is now:
function () {
return toUpperCase.call(slow_foo.apply(this, arguments));
}
This is exactly what we want. Hence our final code is as follows:
var bind = Function.bind;
var call = Function.call;
var bindable = bind.bind(bind); // bindable(f) === f.bind
var callable = bindable(call); // callable(f) === f.call
var someobj = {x: "Quick."};
slow_foo.times_called = 0;
Function.prototype.pipe = function (f) {
var g = this;
return function () {
return f(g.apply(this, arguments));
};
};
function lazyget(obj, key, lazydflt) {
return obj.hasOwnProperty(key) ? obj[key] : evaluate(lazydflt);
}
function slow_foo() {
slow_foo.times_called++;
return "Sorry for keeping you waiting.";
}
function evaluate(thunk) {
return typeof thunk === "function" ? thunk() : thunk;
}
Then we define the test case:
console.log(slow_foo.times_called);
console.log(lazyget(someobj, "x", slow_foo()));
console.log(slow_foo.times_called);
console.log(lazyget(someobj, "x", slow_foo.pipe(callable("".toUpperCase))));
console.log(slow_foo.times_called);
console.log(lazyget(someobj, "y", slow_foo.pipe(callable("".toUpperCase))));
console.log(slow_foo.times_called);
console.log(lazyget(someobj, "y", "slow_foo().toUpperCase()"));
console.log(slow_foo.times_called);
And the result is as expected:
0
Quick.
1
Quick.
1
SORRY FOR KEEPING YOU WAITING.
2
slow_foo().toUpperCase()
2
Hence as you can see for most cases you never need to create thunks manually. Either lift functions using the function lazy to make them return thunks or compose functions to create new thunks.
If you want delayed execution you should look in to using setTimeout.
setTimeout(function() {
console.log("I'm delayed");
}, 10);
console.log("I'm not delayed");
>I'm not delayed
>I'm delayed
https://developer.mozilla.org/en-US/docs/Web/API/window.setTimeout

How to programmatically set the length of a function

The length property of functions tells how long the 'expected' argument list is:
console.log((function () {}).length); /* 0 */
console.log((function (a) {}).length); /* 1 */
console.log((function (a, b) {}).length); /* 2 etc. */
However, it is a readonly method:
f = function (a) {};
alert(f.length); // 1
f.length = 3;
alert(f.length); // 1
Is there a way to programmatically set that length? The closest I've come so far is to use the Function constructor:
f = new Function("a,b,c", "/* function body here */");
f.length; // 3
However, using Function is essentially the same as eval and we all know how bad that is. What other options do I have here?
It turns out the length property on functions is configurable, which means you can use .defineProperty to change the value of length on a function. Example:
function hi() {}
hi.length === 0; // Expected
Object.defineProperty(hi, "length", { value: 5 })
hi.length === 5; // Intriguing
This works in the latest version of Chrome and Firefox, but it does not work in Safari (v9.1.1).
For now, here's the best solution I could think of.
makeFunc = function (length, fn) {
switch (length) {
case 0 : return function () { return fn.apply(this, arguments); };
case 1 : return function (a) { return fn.apply(this, arguments); };
case 2 : return function (a,b) { return fn.apply(this, arguments); };
case 3 : return function (a,b,c) { return fn.apply(this, arguments); };
case 4 : return function (a,b,c,d) { return fn.apply(this, arguments); };
case 5 : return function (a,b,c,d,e) { return fn.apply(this, arguments); };
case 6 : return function (a,b,c,d,e,f) { return fn.apply(this, arguments); };
case 7 : return function (a,b,c,d,e,f,g) { return fn.apply(this, arguments); };
case 8 : return function (a,b,c,d,e,f,g,h) { return fn.apply(this, arguments); };
case 9 : return function (a,b,c,d,e,f,g,h,i) { return fn.apply(this, arguments); };
default : return function (a,b,c,d,e,f,g,h,i,j) { return fn.apply(this, arguments); };
}
};
Example usage:
var realFn = function () {
return "blah";
};
lengthSix = makeFunc(6, realFn);
lengthSix.length; // 6
lengthSix(); // "blah"
Personally, I always cringe whenever I use copy and paste when programming, so I'd be very happy to hear of any better options.
Update
I thought of a method which could work for any arbitrary size, unlike the example above which is limited by how many times you want to copy-and-paste. Essentially, it dynamically creates a function (using new Function) which will return a function of the right size which then just proxies through to whatever function you pass to it. Yeah that does hurt your head. Anyway, I thought I'd benchmark it against the above...
http://jsperf.com/functions-with-custom-length (you can see the 'evil' code there too).
The evil method is many hundreds of times slower than the hacky copypasta method, so there you go.
I am doing something like what you're asking for using roughly the following:
/* Make a new function with a given size */
function SizedFunc(num_args) {
if(SizedFunc.sizedFuncs === undefined) SizedFunc.sizedFuncs = {};
if(SizedFunc.sizedFuncs[num_args] === undefined) {
var argNames = [];
for(var i = 0; i < num_args; ++i) {
argNames.push('arg' + i);
}
SizedFunc.sizedFuncs[num_args] = new Function(argNames, 'return this.apply(null, arguments);');
}
return SizedFunc.sizedFuncs[num_args];
}
This does use a Function constructor but in a strictly limited way and only ever once per function size to create a function wrapper (which is cached for that size) after that I use wrapper.bind(real_function) to provide the implementation as a function expression/object.
Advantages are that any size is supported and we aren't hard coding the function definitions but actual function implementations are never done in an 'eval' like way and the string passed to the Function constructor is always the same.
/* ---- example ---- */
var a = SizedFunc(4).bind(function() { console.log.apply(null, arguments); });
var b = SizedFunc(4).bind(function(a, b, c, d) { console.log(a + b, c + d); });
console.log(typeof a); // -> function
console.log(typeof b); // -> function
console.log(a.length); // -> 4
console.log(b.length); // -> 4
a(1, 2, 3, 4) // -> 1 2 3 4
a(1, 2, 3, 4, 5); // -> 1 2 3 4 5
b(1, 2, 3, 4) // -> 3 7
I'm sure there are plenty of reasons this is bad, too (starting with the use of bind meaning functions created this way can't be bound to an object) but it is useful in situations where one needs to be able to create a function of arbitrary length dynamically.
According to the ECMA Script standard, revision 5.1 on page 103, the .length parameter on a Function object is not writable so it is set when the function is declared and not changable (if implemented per spec).
Thus, the only way to create a function with a particular .length upon demand is to either have a bunch of functions lying around of various length (as nickf suggests), create a Function object (as you've already mentioned) or use eval() with a dynamically created string. I don't know what problem yu're actually trying to solve, but I personally find nothing wrong with using eval() if you know the source of the code you're using it with and have a way of either checking it or knowing what it will or won't have in it. In this case, programmatically generating a certain number of parameters in a string before calling eval() on it poses no security risk I'm aware of.
If this is all your own code, you can just create a new property on the function .dynLength that is mutable and set it to whatever you want and have your code use that.
There’s an npm module util-arity which does what you want:
const arity = require('util-arity');
arity(3, () => {}).length; //» 3

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