For example, I'm writing a function that uses two arrays. How can I define this function to call it as a global function, passing it two arrays or as an Array method, passing it a second array:
func([1, 2, 3], [3, 2, 1]);
[1, 2, 3].func([3, 2, 1]);
Here is my idea:
func = function (...args) { // define global function
if (args.length > 1) {
// function called as global with 2 arrays in arguments
} else {
// function called as Array method, with second array in arguments
}
}
Array.prototype.func = func; // define Array method
Is there a more better way?
Thanks.
You could perhaps create a wrapper function for the prototype:
func = function (a,b) { // define global function
// will always have a,b defined
}
Array.prototype.func = function (b){ func(this,b) }; // define Array method
This way it doesnt make any difference for your function.
(Assuming your function expects exactly 2 arguments)
Related
Can I call a function with array of arguments in a convenient way in JavaScript?
Example:
var fn = function() {
console.log(arguments);
}
var args = [1,2,3];
fn(args);
I need arguments to be [1,2,3], just like my array.
Since the introduction of ES6, you can sue the spread syntax in the function call:
const args = [1,2,3];
fn(...args);
function fn() {
console.log(arguments);
}
Before ES6, you needed to use apply.
var args = [1,2,3];
fn.apply(null, args);
function fn() {
console.log(arguments);
}
Both will produce the equivalent function call:
fn(1,2,3);
Notice that I used null as the first argument of the apply example, which will set the this keyword to the global object (window) inside fn or undefined under strict mode.
Also, you should know that the arguments object is not an array, it's an array-like object, that contains numeric indexes corresponding to the arguments that were used to call your function, a length property that gives you the number of arguments used.
In ES6, if you want to access a variable number of arguments as an array, you can also use the rest syntax in the function parameter list:
function fn(...args) {
args.forEach(arg => console.log(arg))
}
fn(1,2,3)
Before ES6, if you wanted to make an array from your arguments object, you commonly used the Array.prototype.slice method.
function fn() {
var args = Array.prototype.slice.call(arguments);
console.log(args);
}
fn(1,2,3);
Edit: In response to your comment, yes, you could use the shift method and set its returned value as the context (the this keyword) on your function:
fn.apply(args.shift(), args);
But remember that shift will remove the first element from the original array, and your function will be called without that first argument.
If you still need to call your function with all your other arguments, you can:
fn.apply(args[0], args);
And if you don't want to change the context, you could extract the first argument inside your function:
function fn(firstArg, ...args) {
console.log(args, firstArg);
}
fn(1, 2, 3, 4)
In ES5, that would be a little more verbose.
function fn() {
var args = Array.prototype.slice.call(arguments),
firstArg = args.shift();
console.log(args, firstArg);
}
fn(1, 2, 3, 4);
In ECMAScript 6, you can use spread syntax (...) for that purpose. It's way simpler and easier to understand than Function.prototype.apply().
Code example:
const fn = function() {
console.log(arguments);
}
const args = [1,2,3];
fn(...args);
Reading about functional programming - got to currying, example has a simple currying function. I understand everything except the last else block.
var curry = function (fn, fnLength) {
fnLength = fnLength || fn.length;
return function () {
var suppliedArgs = Array.prototype.slice.call(arguments);
if (suppliedArgs.length >= fn.length) {
return fn.apply(this, suppliedArgs);
} else if (!suppliedArgs.length) {
return fn;
} else {
return curry(fn.bind.apply(fn, [this].concat(suppliedArgs)), fnLength - suppliedArgs.length);
}
};
};
If the supplied args are >=, call the function with the supplied arguments.
Else if suppliedArgs.length is falsy, return the original function without doing anything.
Else ???
I think recursively call the function?
I don't understand what .bind.apply accomplishes.
Is [this] just in an array because suppliedArgs.push wouldn't return the array?
Start by looking at how you call Function#bind():
fun.bind(thisArg[, arg1[, arg2[, ...]]])
Then consider how you use Function#apply():
fun.apply(thisArg, [argsArray])
So given for bind() we need to call it on a function, and give it multiple parameters (not an array), and all we have is an array of arguments (suppliedArgs in the code), then how can we do that? Well, the main way you can call a function that takes multiple arguments instead of a single argument that is an array is to use .apply() on the function. So then we have fn.bind.apply(...something...).
The first parameter to .apply() is the this value - which needs to be the function to be bound in the case of .bind() (see below for an explanation of why). Hence fn.bind.apply(fn, ...).
Then, the second parameter to .apply() is an array of all the arguments to the function you are invoking, which in the case of .bind() is thisArg[, arg1[, arg2[, ...]]]. Hence we need a single array, with the first value being the value for this within the function, followed by the other arguments. Which is what [this].concat(suppliedArgs) produces.
So the whole fn.apply.bind(fn, [this].concat(suppliedArgs)) thing produces a correctly bound function that will have the supplied arguments to the current function "prefilled", with the correct this context. This function that is produced is then passed as the fn parameter in a recursive call to curry(), which in turn will produce another function in the same way as the top level call will.
The overall effect is that whenever you call a function created by curry(), if you don't pass the expected number of parameters, you will get a new function which takes the remaining number of parameters, or you will evaluate the original function with the entire list of parameters passed in correctly.
e.g.
function addAllNums(a, b, c, d, e) {
return a + b + c + d + e;
}
var curriedAddAll = curry(addAllNums, 5);
var f1 = curriedAddAll(1); // produces a function expecting 4 arguments
var f2 = f1(2, 3); // f2 is a function that expects 2 arguments
var f3 = f2(4); // f3 is a function that expects 1 argument
var f4 = f3(5); // f4 doesn't expect any arguments
var ans = f4(); // ans = 1 + 2 + 3 + 4 + 5 = 15.
// OR var ans = f3(5); => same result
Why the different thisArg values?
Probably the most confusing thing about this line of code is the two different values for thisArg in .bind() and .apply().
For .apply(), the thisArg is what you want the value of this to be inside the function you are calling .apply() on. e.g. myFunction.apply(myObj, ['param1', 'param2']) is equivalent to myObj.myFunction('param1', 'param2').
In this particular case, .bind() is executed on the fn function, so we want fn to be the this value for .bind(), so it knows what function it is creating a bound version of.
For .bind(), the thisArg is what the value of this will be inside the bound function that is returned.
In our case, we want to return a bound function that has the same this value as we currently have. In other words, we want to maintain the this value correctly within the new function, so it doesn't get lost as you create new functions which happens when you call a curried function with less arguments than it expects.
If we did not maintain the this value correctly, the following example wouldn't log the correct value of this. But by maintaining it, the correct value will be output.
var myObj = {
a: 1,
b: curry(function (a, b, c, d) {
console.log('this = ', this);
return a + b + c + d;
})
};
var c = myObj.b(1,1,1); // c is a function expecting 1 argument
c(1); // returns 4, and correctly logs "this = Object {a: 1, b: function}"
// if "this" wasn't maintained, it would log the value of "this" as
// the global window object.
The last else block is the main and most important part of the curry function, as it is the actual line that carries the logic for currying.
return curry(fn.bind.apply(fn, [this].concat(suppliedArgs)), fnLength - suppliedArgs.length);
This is what returns the new function that needs n-1 arguments from your previous function. Why? It's a combination of multiple things:
fn.bind.apply simply calls a function in the context of the function itself, while supplying the needed args (suppliedArgs). Note how the next parameter to curry is fnLength - suppliedArgs.length, which reduces the arguments needed to what was passed.
Let's explain it with the help of ES6. Things are going to become more obvious.
// Imagine we have the following code written in ES5
function fn(a, b, c) {
console.log(a, b, c);
}
var arr = [1, 2, 3];
var funcWithBoundArguments = fn.bind.apply(fn, [null].concat(arr));
Let's convert ES5 to ES6 code
// ES6
function fn(a, b, c) { console.log(a, b, c) }
let arr = [1,2,3];
let funcWithBoundArguments = fn.bind(null, ...arr)
You see? When you bind a function we have to explicitly enumerate all the arguments like:
fn.bind(null, 1, 2, 3)
But how could we bind the content of an array if we don't know its content in advance?
Right, we have to use .bind.apply() where:
the 1st argument of apply is the function (fn) we bind
the 2nd argument is an array which gets the context (as the first item of array) that we bind our function to and the rest of the items of the array are the arguments (which number is variable) we bind to our function (fn).
When I define a function like this:
var town = 'Liverpool';
foo = function(x,y){
alert(x);
alert(y);
alert(this.town)
}.bind(this);
I have learned that one way to pass the "this" into a function is to use .bind(this) if I'm using browsers IE9 and above. However if I do this then does it mean I need to change the way I define the x and y parameters of my function?
You can call the function just like foo(argument1, argument2)
The bind() method creates a new function that, when called, has its this keyword set to the provided value, with a given sequence of arguments preceding any provided when the new function is called.
You can pass multiple arguments to bind() to make a function with pre-specified initial arguments. These arguments (if any) follow the provided this value and are then inserted at the start of the arguments passed to the target function, followed by the arguments passed to the bound function, whenever the bound function is called.
function list() {
return Array.prototype.slice.call(arguments);
}
var list1 = list(1, 2, 3); // [1, 2, 3]
// Create a function with a preset leading argument
var leadingThirtysevenList = list.bind(undefined, 37);
var list2 = leadingThirtysevenList(); // [37]
var list3 = leadingThirtysevenList(1, 2, 3); // [37, 1, 2, 3]
for further reading
MDN link
I wanted to have an optional boolean parameter to a function call:
function test() {
if (typeof(arguments[0]) === 'boolean') {
// do some stuff
}
// rest of function
}
I want the rest of the function to only see the arguments array without the optional boolean parameter. First thing I realized is the arguments array isn't an array! It seems to be a standard Object with properties of 0, 1, 2, etc. So I couldn't do:
function test() {
if (typeof(arguments[0]) === 'boolean') {
var optionalParameter = arguments.shift();
I get an error that shift() doesn't exist. So is there an easy way to remove an argument from the beginning of an arguments object?
arguments is not an array, it is an array like object. You can call the array function in arguments by accessing the Array.prototype and then invoke it by passing the argument as its execution context using .apply()
Try
var optionalParameter = Array.prototype.shift.apply(arguments);
Demo
function test() {
var optionalParameter;
if (typeof (arguments[0]) === 'boolean') {
optionalParameter = Array.prototype.shift.apply(arguments);
}
console.log(optionalParameter, arguments)
}
test(1, 2, 3);
test(false, 1, 2, 3);
another version I've seen in some places is
var optionalParameter = [].shift.apply(arguments);
Demo
function test() {
var optionalParameter;
if (typeof (arguments[0]) === 'boolean') {
optionalParameter = [].shift.apply(arguments);
}
console.log(optionalParameter, arguments)
}
test(1, 2, 3);
test(false, 1, 2, 3);
As Arun pointed out arguments is not an array
You will have to convert in into an array
var optionalParameter = [].shift.apply(arguments);
It's not fancy but the best solution to remove the first argument without side effect (without ending with an additional argument as would do shift) would probably be
for (var i=0;i<arguments.length;i++) arguments[i]=arguments[i+1];
Example :
function f(a, b, c, d) {
for (var i=0;i<arguments.length;i++) arguments[i]=arguments[i+1];
console.log(a,b,c,d);
}
f(1,2,3,4); // logs 2,3,4,undefined
I just started reading Functional JavaScript and immediately was introduced to a function that I don't understand:
function splat(fun) {
return function(array) {
return fun.apply(null, array);
};
}
var addArrayElements = splat(function(x, y) { return x + y });
addArrayElements([1, 2]);
//=> 3
How does splat(function(x, y) { return x + y }) work. It's called with the array [1,2], but it seems like the anonymous function inside the call to splat takes two parameters, not one array.
Putting console.log(fun) on line 2 of this code shows that fun is the entirety of the anonymous function(x, y) { return x + y }. console.log(array) after return function(array) { shows that array is [1, 2]. Where does array come from then?
Thanks much.
It might be simpler to see how this function would have been written without using the .apply method:
function splat(fun) {
return function(array) {
return fun(array[0], array[1]);
};
}
First you call splat, passing it a function:
var add = function(x,y){ return x + 1 };
var ff = splat(add);
At this point, ff refers to the function(array) function, meaning its an one-argument function. The private variable fun refers to the add function.
Now, you call ff passing its one argument
ff([1,2]);
and it uses the values in the array to call fun with two arguments
return fun(array[0], array[1]);
The only difference between this and the real example is that the apply method lets you work with any argument array length instead of hardcoding a specific length (2) like I did.
//Every time we call this function, we get another one back
function splat(fun) {
return function(array) { // <-- this one will be returned in splat();
return fun.apply(null, array);
};
}
//Step one, call splat, pass a function as parameter
var addArrayElements = splat(function(x, y) { return x + y });
/*
Get back a function that accepts an array, and will execute the function we just passed in on it
*/
// This will call the newly created function, func will be available because it's in a closure
addArrayElements([1, 2]);
The last thing is that, even if the anonymous function takes two parameters, we call apply on it so it will bind array[0] ==> x and array[1] ==> y
This is an example of a higher order function. That's a function that takes functions as arguments and returns functions instead of just regular values (though functions are "just regular values" in Javascript). In this case:
function splat(fun) {
splat takes a function as its argument...
return function(array) {
...and returns a new function which takes an array...
return fun.apply(null, array);
...and when called calls the first fun function with the array .applied as its arguments.
So splat takes one function which expects several parameters and wraps it in a function which takes an array of parameters instead. The name "splat" comes from languages like Ruby, where a * (a "splat" or "squashed bug") in the parameter list of a function accumulates an arbitrary number of arguments into an array.
var addArrayElements = splat(function(x, y) { return x + y });
addArrayElements is now basically:
function (array) {
// closed over variable:
// var fun = function(x, y) { return x + y }
return fun.apply(null, array);
}
Here this is realized by a closure, which closes over and "preserves" the original fun passed to splat in the new returned function.
addArrayElements = function(array) { fun.apply(null, array); };
BUT
it has a closure whereby the variable context of its containing scope (that of the splat function that created the anonymous function) remains visible and accessible.
In JavaScript, functions are first-class objects that can be referenced and passed as arguments or, as in this case, through the closure mechanism.
Edit: about JavaScript and scope
In most languages, variables are, by default, local to the scope they're defined in (which usually is a function's local symbol table). By contrast, in JavaScript a variable is local only if it is defined using the var keyword; otherwise, the symbol will be looked back in the chain of the containing scopes, up to the implicit root object (which in the case of web browsers is window. I.e.,
function foo() { someVar = "bar"; }
foo();
alert(someVar); // shows "bar"
Not being restricted to the local scope, the symbol has been (purposely or not) leaked to the root scope.
Taking it one step further:
function foo() {
var baz = function() {
someVar = "bar";
};
baz();
}
foo();
alert(someVar); // shows "bar"
However, if you declare someVar within foo:
function foo() {
var someVar;
var baz = function() {
someVar = "bar";
};
baz();
alert("someVar in foo=" + someVar); // shows "bar"
}
foo();
alert("someVar in root=" + window.someVar); // shows "undefined"
Note that in this last version I needed to use window.someVar instead of just someVar because someVar never got defined as a variable in the root scope nor as a property of the root object, which caused an error.
a more functional approach uses bind(), which is short enough you don't really need splat anymore, and it's always nice to eliminate closures:
var addArrayElements = Function.apply.bind( function(x, y) { return x + y } , null );
addArrayElements([1, 2]); // === 3