How can one define a pair of functions that call each other in Javascript so that JS-lint does not complain that 'factorial' is being used before it is defined?
function factorial1(n) { return factorial(n); }
function factorial(n) { return n === 0 ? 1 : n * factorial1(n - 1); }
It seems there is no valid ordering that will satisfy JSlint.
(One can be embedded in the other, but this would be a nightmare for a collection of functions that all call each other).
Surely this is handled by the language right?
Is this just a bug in JSlint?
(This question must have been answered somewhere, but I cannot find it!)
The references inside the functions is not resolved until they are executed. As long as both functions has been defined by the time one of them executes, they will find each other.
If you want to get rid of the JSLint warning, you could define the name of the function, just before:
var factorial;
function factorial1(n) { return factorial(n); }
function factorial(n) { return n === 0 ? 1 : n * factorial1(n - 1); }
I prefer this syntax to functions:
var factorial1 = function(n) { return factorial(n); }
var factorial = function(n) { return n === 0 ? 1 : n * factorial1(n - 1); }
Will not be able to forget to "declare" function with this syntax.
Another approach to this problem, is to pass it as a callback function, like this:
const factorial1 = (n, callback) => { return callback(n); }
const factorial = (n) => { return n === 0 ? 1 : n * factorial1(n - 1, factorial ); }
Related
const Calculate = {
factorial(n) {
return n * factorial(n - 1)
}
}
let newNumber = Calculate.factorial(8);
So every time I call the function I get a factorial not defined error. I'm guessing it has something to do with the function being inside an object. Need some help in understanding whats going on. Thank you in advance
Objects don't create a variable scope, so you need to refer to the object when calling the method.
You can also use this to refer to the object that the method was called on, so you don't have to hard-code the variable name.
const Calculate = {
factorial(n) {
if (n <= 1) {
return 1;
}
return n * this.factorial(n - 1)
}
}
let newNumber = Calculate.factorial(8);
console.log(newNumber);
I have an arrow function, offered graciously by Ele of the community here, but for the life of me, I can't understand it:
let isValid = function (arr, arr2) {
let sum = (array, n) => array.reduce((a, an) => a + (an === n), 0);
return !arr2.some(n => !arr.some(an => an === n && sum(arr, an) === sum(arr2, n)))
};
Would someone be so kind as to translate this into a standard function so that I can follow it on my skill level?
Thank you.
My assumption:
function isValid (arr, arr2) {
...this is where i'm lost
}
Your assumption is correct for the outer function. The first line inside it would become:
function sum(array, n) {
return array.reduce(function(a, an) {
return a + (an === n);
}, 0);
Have a read about Arrow Functions and how they differ from conventional function declarations. Mostly (but not completely), they're just syntactical sugar vs. conventional functions.
They differ most markedly in terms of context, i.e. what this points to inside the function body. Arrow functions always run in the outer, prevailing context in which the function was declared. Conventional functions, e.g. via bind(), can be repointed to a different context.
let foo = function() {
let bar = () => this;
return bar();
}.bind('a');
foo(); //'a', because foo()'s context is 'a'
So how about that sugar? It can look confusing at first, especially when you have multiple arrow functions in one line. One key thing to remember is that they're just implicit shorthand for things you used to have to code manually.
let foo = a => a+1;
Is the same as
function foo2(a) { return a + 1; }
(The hoisting will be different, but that's a bit beyond the scope of this answer.)
One thing we can tell from the above is there is that, where the part after => is a single statement, it's interpreted as a return value, without us having to actually write return.
foo(1); //2
This is great for simple functions that do one job expressable in one line of code. If you need more verbose functions you enclose the code in {} as usual.
let foo3 = a => {
return a+1;
};
This again works identically to the foo and foo2 above.
And so, finally, breaking down that fearsome line:
let sum = (array, n) => array.reduce((a, an) => a + (an === n), 0);
It says:
assign a function to the local scope variable sum
it accepts two arguments, array and n
it has one job to do, expressable as one line of code and so no need for binding { and } for the function body. That job is to call reduce() and return (implicitly) the value
The first argument to reduce, the callback, accepts two arguemnts, a and an
This callback, like the one to reduce, also has just one job to do, and that job is to return the value of a + (an === n)
One final word on the sugar, which you may have spotted above, is that, with arrow functions, if only a single argument is accepted you don't need to wrap it in brackets. Multiple arguments are, though, and comma-separated as usual.
let foo = single_arg => alert(1);
let foo2 = (arg1, arg2) => alert(2);
Hope this helps.
You can use https://babeljs.io/ to compile from this new javscript to "old" javascript. You can try it directly on its home page.
Here's the output it gives:
var isValid = function isValid(arr, arr2) {
var sum = function sum(array, n) {
return array.reduce(function (a, an) {
return a + (an === n);
}, 0);
};
return !arr2.some(function (n) {
return !arr.some(function (an) {
return an === n && sum(arr, an) === sum(arr2, n);
});
});
};
That approach uses a lot of arrow functions, and can be translated to the following standard function declarations:
let isValid = function(arr, arr2) {
let sum = function (array, n) {
return array.reduce(function(a, an) {
return a + (an === n); // coercion -> true = 1, false = 0
}, 0);
};
return !arr2.some(function(n) {
let sum2 = sum(arr2, n);
return !arr.some(function(an) {
return an === n && sum(arr, an) === sum2;
});
});
};
I'm trying to solve a puzzle, and am at my wit's end trying to figure it out.
I'm supposed to make a function that works like this:
add(1); //returns 1
add(1)(1); //returns 2
add(1)(1)(1); //returns 3
I know it can be done because other people have successfully completed the puzzle. I have tried several different ways to do it. This is my most recent attempt:
function add(n) {
//Return new add(n) on first call
if (!(this instanceof add)) {
return new add(n);
}
//Define calc function
var obj = this;
obj.calc = function(n) {
if (typeof n != "undefined") {
obj.sum += n;
return obj.calc;
}
return obj.sum;
}
//Constructor initializes sum and returns calc(n)
obj.sum = 0;
return obj.calc(n);
}
The idea is that on the first call, a new add(n) is initialized and calc(n) is run. If calc receives a parameter, it adds n to sum and returns itself. When it eventually doesn't receive a parameter, it returns the value of sum.
It makes sense in theory, but I can't get it to work. Any ideas?
--edit--
My code is just the route I chose to go. I'm not opposed to a different approach if anyone can think of one.
To answer "how dow this work". Given:
function add(n) {
function calc(x) {
return add(n + x);
}
calc.valueOf = function() {
return n;
}
return calc;
}
var sum = add(1)(2)(3); // 6
When add is called the first time, it stores the value passed in in a variable called n. It then returns the function calc, which has a closure to n and a special valueOf method (explained later).
This function is then called with a value of 2, so it calls add with the sum of n + x, wich is 1 + 2 which 3.
So a new version of calc is returned, this time with a closure to n with a value of 3.
This new calc is called with a value of 3, so it calls add with n + x, which this time is 3 + 3 which is 6
Again add returns a new calc with n set to 6. This last time, calc isn't called again. The returned value is assigned to the variable sum. All of the calc functions have a special valueOf method that replaces the standard one provided by Object.prototype. Normally valueOf would just return the function object, but in this case it will return the value of n.
Now sum can be used in expressions, and if its valueOf method is called it will return 6 (i.e. the value of n held in a closure).
This seems pretty cool, and sum will act a lot like a primitve number, but it's actually a function:
typeof sum == 'function';
So be careful with being strict about testing the type of things:
sum * 2 // 12
sum == 6 // true
sum === 6 // false -- oops!!
Here's a somewhat streamlined version of #RobG's great answer:
function add(n) {
function calc(x) { return n+=x, calc; }
calc.valueOf = function() { return n; };
return calc;
}
The minor difference is that here calc just updates n and then returns itself, rather than returning itself via another call to add, which puts another frame on the stack.
Making self-replication explicit
calc is thus a pure self-replicating function, returning itself. We can encapsulate the notion of "self replication" with the function
function self_replicate(fn) {
return function x() {
fn.apply(this, arguments);
return x;
};
}
Then add could be written in a possibly more self-documenting way as
function add(n) {
function update(x) { n += x; }
var calc = self_replicate(update);
calc.valueOf = function() { return n; };
return calc;
}
Parallel to Array#reduce
Note that there is a certain parallelity between this approach to repeatedly calling a function and Array#reduce. Both are reducing a list of things to a single value. In the case of Array#reduce the list is an array; in our case the list is parameters on repeated calls. Array#reduce defines a standard signature for reducer functions, namely
function(prev, cur)
where prev is the "accumulator" (value so far), cur is the new value being fed in, and the return value becomes the new value the accumulator. It seems useful to rewrite our implementation to make use of a function with that kind of signature:
function add(n) {
function reducer(prev, cur) { return prev + cur; }
function update(x) { n = reducer(n, x); }
var calc = self_replicate(update);
calc.valueOf = function() { return n; };
return calc;
}
Now we can create a more general way to create self-replication-based reducers based on a reducer function:
function make_repeatedly_callable_function(reducer) {
return function(n) {
function update(x) { n = reducer(n, x); }
var calc = self_replicate(update);
calc.valueOf = function() { return n; };
return calc;
};
}
Now we can create add as
var add = make_repeatedly_callable_function(function(prev, cur) { return prev + cur; });
add(1)(2);
Actually, Array#reduce calls the reducer function with third and fourth arguments, namely the index into the array and the array itself. The latter has no meaning here, but it's conceivable we might want something like the third argument to know what "iteration" we're on, which is easy enough to do by just keeping track using a variable i:
function reduce_by_calling_repeatedly(reducer) {
var i = 0;
return function(n) {
function update(x) { n = reducer( n, x, i++); }
var calc = self_replicate(update);
calc.valueOf = function() { return n; };
return calc;
};
}
Alternative approach: keeping track of values
There are certain advantages to keeping track of the intermediate parameters the function is being called with (using an array), and then doing the reduce at the end instead of as we go along. For instance, then we could do Array#reduceRight type things:
function reduce_right_by_calling_repeatedly(reducer, initialValue) {
var array_proto = Array.prototype,
push = array_proto.push,
reduceRight = array_proto.reduceRight;
return function(n) {
var stack=[],
calc = self_replicate(push.bind(stack));
calc.valueOf = reduceRight.bind(stack, reducer, initialValue);
return calc(n);
};
}
Non-primitive objects
Let's try using this approach to build ("extend") objects:
function extend_reducer(prev, cur) {
for (i in cur) {
prev[i] = cur[i];
}
return prev;
}
var extend = reduce_by_calling_repeatedly(extend_reducer);
extend({a: 1})({b: 2})
Unfortunately, this won't work because Object#toValue is invoked only when JS needs a primitive object. So in this case we need to call toValue explicitly:
extend({a: 1})({b: 2}).toValue()
Thanks for the tip on valueOf(). This is what works:
function add(n) {
var calc = function(x) {
return add(n + x);
}
calc.valueOf = function() {
return n;
}
return calc;
}
--edit--
Could you please explain how this works? Thanks!
I don't know if I know the correct vocabulary to describe exactly how it works, but I'll attempt to:
Example statement: add(1)(1)
When add(1) is called, a reference to calc is returned.
calc understands what n is because, in the "mind" of the interpreter, calc is a function child of add. When calc looks for n and doesn't find it locally, it searches up the scope chain and finds n.
So when calc(1) is called, it returns add(n + x). Remember, calc knows what n is, and x is simply the current argument (1). The addition is actually done inside of calc, so it returns add(2) at this point, which in turn returns another reference to calc.
Step 2 can repeats every time we have another argument (i.e. (x)).
When there aren't any arguments left, we are left with just a definition of calc. The last calc is never actually called, because you need a () to call a function. At this point, normally the interpreter would return a the function object of calc. But since I overrode calc.valueOf it runs that function instead.
When calc.valueOf runs, it finds the most recent instance of n in the scope chain, which is the cumulative value of all previous n's.
I hope that made some sense. I just saw #RobG 's explanation, which is admittedly much better than mine. Read that one if you're confused.
Here's a variation using bind:
var add = function _add(a, b) {
var boundAdd = _add.bind(null, a + b);
boundAdd.valueOf = function() {
return a + b;
}
return boundAdd;
}.bind(null, 0);
We're taking advantage of a feature of bind that lets us set default arguments on the function we're binding to. From the docs:
bind() also accepts leading default arguments to provide to the target
function when the bound function is called.
So, _add acts as a sort of master function which takes two parameters a and b. It returns a new function boundAdd which is created by binding the original _add function's a parameter to a + b; it also has an overridden valueOf function which returns a + b (the valueOf function was explained quite well in #RobG's answer).
To get the initial add function, we bind _add's a parameter to 0.
Then, when add(1) is called, a = 0 (from our initial bind call) and b = 1 (passed argument). It returns a new function where a = 1 (bound to a + b).
If we then call that function with (2), that will set b = 2 and it'll return a new function where a = 3.
If we then call that function with (3), that will set b = 3 and it'll return a new function where a = 6.
And so on until valueOf is called, at which point it'll return a + b. Which, after add(1)(2)(3), would be 3 + 3.
This is a very simple approach and it meets the criteria the OP was looking for. Namely, the function is passed an integer, keeps track of that integer, and returns itself as a function. If a parameter is not passed - the function returns the sum of the integers passed to it.
let intArray = [];
function add(int){
if(!int){
return intArray.reduce((prev, curr) => prev + curr)
}
intArray.push(int)
return add
}
If you call this like so:
console.log(add(1)(1)());
it outputs 2.
I trying to make next with closure:
function func(number) {
var result = number;
var res = function(num) {
return result + num;
};
return res;
}
var result = func(2)(3)(4)(5)(3);
console.log(result); // 17
I need to receive 2 + 3 + 4 + 5 + 3 = 17
But I got an error: Uncaught TypeError: number is not a function
You somehow have to signalize the end of the chain, where you are going to return the result number instead of another function. You have the choice:
make it return a function for a fixed number of times - this is the only way to use the syntax like you have it, but it's boring. Look at #PaulS' answer for that. You might make the first invocation (func(n)) provide the number for how many arguments sum is curried.
return the result under certain circumstances, like when the function is called with no arguments (#PaulS' second implementation) or with a special value (null in #AmoghTalpallikar's answer).
create a method on the function object that returns the value. valueOf() is suited well because it will be invoked when the function is casted to a primitive value. See it in action:
function func(x) {
function ret(y) {
return func(x+y);
}
ret.valueOf = function() {
return x;
};
return ret;
}
func(2) // Function
func(2).valueOf() // 2
func(2)(3) // Function
func(2)(3).valueOf() // 5
func(2)(3)(4)(5)(3) // Function
func(2)(3)(4)(5)(3)+0 // 17
You're misusing your functions.
func(2) returns the res function.
Calling that function with (3) returns the number 5 (via return result + num).
5 is not a function, so (4) gives an error.
Well, the (2)(3) part is correct. Calling func(2) is going to return you res, which is a function. But then, calling (3) is going to return you the result of res, which is a number. So the problem comes when you try to call (4).
For what you're trying to do, I don't see how Javascript would predict that you're at the end of the chain, and decide to return a number instead of a function. Maybe you could somehow return a function that has a "result" property using object properties, but mostly I'm just curious about why you're trying to do things this way. Obviously, for your specific example, the easiest way would just be adding the numbers together, but I'm guessing you're going a bit further with something.
If you want to keep invoking it, you need to keep returning a function until you want your answer. e.g. for 5 invocations
function func(number) {
var result = number,
iteration = 0,
fn = function (num) {
result += num;
if (++iteration < 4) return fn;
return result;
};
return fn;
}
func(2)(3)(4)(5)(3); // 17
You could also do something for more lengths that works like this
function func(number) {
var result = number,
fn = function () {
var i;
for (i = 0; i < arguments.length; ++i)
result += arguments[i];
if (i !== 0) return fn;
return result;
};
return fn;
}
func(2)(3, 4, 5)(3)(); // 17
I flagged this as a duplicate, but since this alternative is also missing from that question I'll add it here. If I understand correctly why you would think this is interesting (having an arbitrary function that is applied sequentially to a list of values, accumulating the result), you should also look into reduce:
function sum(a, b) {
return a + b;
}
a = [2, 3, 4, 5, 3];
b = a.reduce(sum);
Another solution could be just calling the function without params in order to get the result but if you call it with params it adds to the sum.
function add() {
var sum = 0;
var closure = function() {
sum = Array.prototype.slice.call(arguments).reduce(function(total, num) {
return total + num;
}, sum);
return arguments.length ? closure : sum;
};
return closure.apply(null, arguments);
}
console.log(add(1, 2, 7)(5)(4)(2, 3)(3.14, 2.86)); // function(){}
console.log(add(1, 2, 7)(5)(4)(2, 3)(3.14, 2.86)()); // 30;
We can make light work of it using a couple helper functions identity and sumk.
sumk uses a continuation to keep a stack of the pending add computations and unwinds the stack with 0 whenever the first () is called.
const identity = x => x
const sumk = (x,k) =>
x === undefined ? k(0) : y => sumk(y, next => k(x + next))
const sum = x => sumk(x, identity)
console.log(sum()) // 0
console.log(sum(1)()) // 1
console.log(sum(1)(2)()) // 3
console.log(sum(1)(2)(3)()) // 6
console.log(sum(1)(2)(3)(4)()) // 10
console.log(sum(1)(2)(3)(4)(5)()) // 15
I trying to make next with closure:
function func(number) {
var result = number;
var res = function(num) {
return result + num;
};
return res;
}
var result = func(2)(3)(4)(5)(3);
console.log(result); // 17
I need to receive 2 + 3 + 4 + 5 + 3 = 17
But I got an error: Uncaught TypeError: number is not a function
You somehow have to signalize the end of the chain, where you are going to return the result number instead of another function. You have the choice:
make it return a function for a fixed number of times - this is the only way to use the syntax like you have it, but it's boring. Look at #PaulS' answer for that. You might make the first invocation (func(n)) provide the number for how many arguments sum is curried.
return the result under certain circumstances, like when the function is called with no arguments (#PaulS' second implementation) or with a special value (null in #AmoghTalpallikar's answer).
create a method on the function object that returns the value. valueOf() is suited well because it will be invoked when the function is casted to a primitive value. See it in action:
function func(x) {
function ret(y) {
return func(x+y);
}
ret.valueOf = function() {
return x;
};
return ret;
}
func(2) // Function
func(2).valueOf() // 2
func(2)(3) // Function
func(2)(3).valueOf() // 5
func(2)(3)(4)(5)(3) // Function
func(2)(3)(4)(5)(3)+0 // 17
You're misusing your functions.
func(2) returns the res function.
Calling that function with (3) returns the number 5 (via return result + num).
5 is not a function, so (4) gives an error.
Well, the (2)(3) part is correct. Calling func(2) is going to return you res, which is a function. But then, calling (3) is going to return you the result of res, which is a number. So the problem comes when you try to call (4).
For what you're trying to do, I don't see how Javascript would predict that you're at the end of the chain, and decide to return a number instead of a function. Maybe you could somehow return a function that has a "result" property using object properties, but mostly I'm just curious about why you're trying to do things this way. Obviously, for your specific example, the easiest way would just be adding the numbers together, but I'm guessing you're going a bit further with something.
If you want to keep invoking it, you need to keep returning a function until you want your answer. e.g. for 5 invocations
function func(number) {
var result = number,
iteration = 0,
fn = function (num) {
result += num;
if (++iteration < 4) return fn;
return result;
};
return fn;
}
func(2)(3)(4)(5)(3); // 17
You could also do something for more lengths that works like this
function func(number) {
var result = number,
fn = function () {
var i;
for (i = 0; i < arguments.length; ++i)
result += arguments[i];
if (i !== 0) return fn;
return result;
};
return fn;
}
func(2)(3, 4, 5)(3)(); // 17
I flagged this as a duplicate, but since this alternative is also missing from that question I'll add it here. If I understand correctly why you would think this is interesting (having an arbitrary function that is applied sequentially to a list of values, accumulating the result), you should also look into reduce:
function sum(a, b) {
return a + b;
}
a = [2, 3, 4, 5, 3];
b = a.reduce(sum);
Another solution could be just calling the function without params in order to get the result but if you call it with params it adds to the sum.
function add() {
var sum = 0;
var closure = function() {
sum = Array.prototype.slice.call(arguments).reduce(function(total, num) {
return total + num;
}, sum);
return arguments.length ? closure : sum;
};
return closure.apply(null, arguments);
}
console.log(add(1, 2, 7)(5)(4)(2, 3)(3.14, 2.86)); // function(){}
console.log(add(1, 2, 7)(5)(4)(2, 3)(3.14, 2.86)()); // 30;
We can make light work of it using a couple helper functions identity and sumk.
sumk uses a continuation to keep a stack of the pending add computations and unwinds the stack with 0 whenever the first () is called.
const identity = x => x
const sumk = (x,k) =>
x === undefined ? k(0) : y => sumk(y, next => k(x + next))
const sum = x => sumk(x, identity)
console.log(sum()) // 0
console.log(sum(1)()) // 1
console.log(sum(1)(2)()) // 3
console.log(sum(1)(2)(3)()) // 6
console.log(sum(1)(2)(3)(4)()) // 10
console.log(sum(1)(2)(3)(4)(5)()) // 15