I have two javascript closures and I'm trying to understand why one will accept and input with a particular syntax and the other will reject.
function multiply(factor) {
var ace = (function(number) {
return number*factor;
});
return ace;
}
var yup = multiply(4);
console.log(yup(5));
This outputs 20 to the console as it should.
The second Closure I have is
var k = 3;
var add = (function () {
console.log(k);
var counter = k;
return function (j) {counter += 1; return counter*j}
})(k);
add();
console.log(add(5));
The output is 20 as it should be.
This issue I'm having that if I try to use the syntax of
(function() {
})(number);
In the first closure it does not work and outputs "number is not defined"
And if I try to input into the second closure
(function (k) {
var counter = k;
return function (j) {counter += 1; return counter*j}
});
I get out
function (j) {counter += 1; return counter*j}
to the console.
My question is, what am I not understanding about closers the () at the end of them.
The difference is whether you are creating the closure right away through an IIFE, or a function that makes the closure when called.
Your first snippet written in the second style would be
var yup = (function multiply(factor) {
return function ace(number) {
return number*factor;
};
})(4); // the multiply(4) call is inlined into the statement with the definition
console.log(yup(5));
Your second snippet written in the first style would be
function makeAdd(k) {
console.log(k);
var counter = k;
return function (j) {
counter += 1;
return counter*j;
}
}
var add = makeAdd(3);
add();
console.log(add(5));
Related
Let's consider I have the following function call,
function add(){
x = 0 ;
for(i = 0 i < ##; i++){ // need to run a loop four times
x+=1
}
}
Let's consider I am trying to Implement the function that will add one on each subsequent call, like below
console.log(add()()().getValue()); // 3
console.log(add().getValue()); // 1
console.log(add()().getValue()); // 2
A call to add must return a function which also has a getValue method, and each call to that function must return the same thing. So:
function add() {
var x = 1;
function inner() {
x += 1;
return inner;
}
inner.getValue = function () {
return x;
}
return inner;
}
console.log(add()()().getValue()); // 3
console.log(add().getValue()); // 1
console.log(add()().getValue()); // 2
My guess is they were expecting you to use toString() which is not the greatest way of doing this.
function add(x = 0) {
function next() {
return add(x+1);
}
next.toString = function () {
return x;
};
return next;
}
console.log("example 1", add()()()());
console.log("example 2", add()()()()()()()()());
I think you are trying to emulate the behavior of generator functions. Here is a snippet that illustrates one way you could do it with a generator.
function* adder() {
let x = 0;
while (true) {
yield x + 1;
x++;
}
}
const add = adder();
const firstValue = add.next();
const secondValue = add.next();
const thirdValue = add.next().value;
Everytime I run this function, the p1_Balance will always reset back to 10 and will not hold the new value of an increment or decrement.
function Balance() {
var p1_Balance=10;
var x= Math.floor(10*Math.random());
if (x<5) {
p1_Balance=p1_Balance-1;
} else {
p1_Balance=p1_Balance+1;
}
return p1_Balance;
}
Pass p1_Balance into the function instead of initializing it each time the function is called with: var p1_Balance = 10;
p1_Balance should be declared outside the scope of the function (meaning not within the function itself). Otherwise, each time the function is called, the initializer that sets the value to 10 runs as well.
var p1_Balance=10;
function Balance(){ ...
You can use Javascript closures to create a function that does what you want, as you can see below:
var Balance = (function() {
var p1_Balance = 10;
return function() {
var x = Math.floor(10 * Math.random());
if (x < 5)
return p1_Balance += 1;
else
return p1_Balance -= 1;
};
})();
for (var i = 0; i < 10; i++)
console.log(Balance());
Alternatively, you will need to define the p1_Balance variable outside the function or pass it as an argument.
There could be several solutions:
one is declaring p1_Balance as a global variable.
var p1_Balance=10;
function Balance(){
var x= Math.floor(10*Math.random());
if (x<5) {
p1_Balance=p1_Balance-1;
}
else {
p1_Balance=p1_Balance+1;
}
return p1_Balance;
}
another is you could pass balance as a function parameter:
function Balance(p1_Balance){
var x= Math.floor(10*Math.random());
if (x<5) {
p1_Balance=p1_Balance-1;
}
else {
p1_Balance=p1_Balance+1;
}
return p1_Balance;
}
.....
value = Balance(10);// value=something that you want to change by that function.
I have a 'twice' function that return 2 of the argument passed into it. I also have another function 'runTwice' that counts the number of times it called the 'twice' function (the idea being that I want the 'twice' function to only run 'twice' no matter how often it is called via the 'runTwice' function). Can you please help?
Functions are given below:
var count = 1;
function twice(num){
return num*2;
}
function runTwice(func){
if (count<3){
count++;
return func;
} else {
return 'Function cannot run!';
}
}
var myFunc = runTwice(twice)
var output = [];
for (var i = 0; i < 3; i++){
output.push(myFunc(i));
}
console.log(output);
I would like the output to be [0, 2, 'Function cannot run!'].
I can make this work if I count the 'twice' function directly but I am looking to understand why this doesn't work as presented above.
Just for fun I'll make a generic expireAfter(invocable[, times[, message]]) function:
function expireAfter(invocable, times = 2, message = 'Function cannot run!') {
return function expires() {
if (times > 0) {
times--;
return invocable.apply(this, arguments);
}
return message;
}
}
function twice(n) {
return n * 2;
}
var myFunc = expireAfter(twice);
console.log(Array(3)
.fill()
.map((_, index) => myFunc(index))
);
The function runTwice should return another function that will decide whether to call the function func (using Function.prototype.apply) or to return a string message instead:
function twice(num){
return num * 2;
}
function runTwice(func){
var count = 0; // this will be trapped in a closure along with func
return function() { // this is the function that gets called
count++; // it increments its version of the count variable
if(count <= 2) // if count is less than 2
return func.apply(this, arguments); // then it calls the function func with whatever arguments passed into it and return the returned value of that call
return "Not available anymore!"; // otherwise (count > 2), then it returns a string
}
}
var myFunc = runTwice(twice);
for (var i = 0; i < 3; i++){
console.log(myFunc(i));
}
Even better:
You can pass in the number of times allowed as well:
function double(num) {
return num * 2;
}
function triple(num) {
return num * 3;
}
function run(func, times){
var count = 0; // this will be trapped in a closure along with func and times
return function() { // this is the function that gets called
count++; // it increments its version of the count variable
if(count <= times) // if count is less than times
return func.apply(this, arguments); // then it calls the function func with whatever arguments passed into it and return the returned value of that call
return "Not available anymore!"; // otherwise (count > times), then it returns a string
}
}
var double2times = run(double, 2); // double2times can only be called 2 times
var triple5times = run(triple, 5); // triple5times can only be called 5 times
for (var i = 0; i < 10; i++){
console.log("Double:", double2times(i));
console.log("Triple:", triple5times(i));
}
Let me propose an example that works, then follow up with what fails, highlighting the point to my question.
Here, we have 3 functions being called (1 named, 2 anonymous):
var add = function(a, b) {return a+b};
var multiply = function(a, b) {return a*b};
function myFunction(fxn) {
return function(x) {
return function(y) {
return fxn(x,y);
}
}
}
myFunction(add)(2)(3)
Understandably, this call fails:
myFunction(add)(2)(3)(4)
How would I detect how many functions are being called? In the 2nd call, I'm calling 4 functions (1 named, 3 anonymous).
How would I rewrite the myFunction function in a way that compensated for any given amount of calls? I know we can detect how many arguments a function was given, but is there a way to detect how many functions are being called? I hope I worded this correctly. Thanks.
To find out if a variable contains a reference to a function you can use below code:
if (typeof(v) === "function") alert("This is a function")
Based on above you can find out on how many nested functions there are
function myFunction() {
return function() {
return function() {
return 1 + 2;
}
}
}
var count = 0;
var v = myFunction();
while (typeof(v) === "function") {
count++;
v = v();
}
alert("Nr of nested functions: " + count)
Even if this has no practical use case I can think of, this is a possible solution:
var add = function(a, b) {
return a + b
};
var multiply = function(a, b) {
return a * b
};
var counter = 0;
var result = 0;
function myFunction(fxn) {
counter = 1;
result = 0;
return function first(x) {
++counter;
return function second(y) {
++counter;
x = result ? result : x;
result = fxn(x, y);
return second;
}
}
}
myFunction(add)(1)(2)(3)(4);
alert('Result is: ' + result + '; Parentheses count: ' + counter);
What is the difference between
settings = {
edit: function (key, value) {
return anotherFunction(key, value) {
return value * 2;
};
}
};
and
settings = {
edit: function edit(key, value) {
return anotherFunction(key, value) {
return value * 2;
};
}
};
?
There's no difference when executing.
However, in the second case (named function), you can call the function recursively easier because it has a name.
For example, with a named function you can do:
fact: function factorial(n) {
if(n == 0) return 1;
return n * factorial(n-1); //You can do this with a named function easily
}
Without a name, this would be tricky.
Cheers
The essential difference is better debugging. In your developer tools, the named function in your second example will appear as edit in a backtrace; your first example will appear as anonymous. This can be extremely confusing when you're 10 function deep, and they are all called anonymous.
There are three reasons to give a function an inherent name. The first is that everyone does it. It's what everyone is used to.
function factorial(n) {
var accum = 1, i;
for (i = 1; i <= n; i++) {
accum *= i;
}
return accum;
}
The second is to understand stack traces better, as #meagar wrote.
The third is to let you write call functions recursively.
var factorial = function(n) {
var a = 1;
return (function factRecursive(k, a) {
if (k >= 2) {return factRecursive(k - 1, k * a)}
else {return a;}
})(n, a);
}