Value is not returned - javascript

Hey guys I am beginner at JavaScript and I faced the confusion with the way how return works in JavaScript, that is, if you look at the code below I want to get value 5 when mirzhal method is called but I get nothing thus I need your help
let mir = {
mirzhal() {
function m() {
let mi = 5;
return alert(mi);
}
return m;
}
}
mir.mirzhal();

You need to return 5 from m function and also need to call the returned function
let mir = {
mirzhal() {
function m() {
let mi = 5;
return mi;
}
return m;
}
}
console.log(mir.mirzhal()());

return will return whatever is on the right-hand side of it.
If you want to get 5 then you need to return 5.
return 5
… and you also need to examine the return value:
const five = mir.mirzhal();
alert(five);
Your current code doesn't return 5. It returns a function.
If you were to call that function, then it would alert 5 (and return the return value of that function).
const a_function = mir.mirzhal();
a_function();

Here are your problems:
in function m, you return the result of alert(mi) which is undefined (= nothing is javascript)
in function mirzhal you return a function, so when you execute mirzhal() your result is a function you need to call.
mir.mirzhal()(); will work
This should be better:
let mir = {
mirzhal() {
return 5;
}
}
mir.mirzhal();

It was because the function mirzhal() returns another function.
Try like this:
mir.mirzhal()();
And you can see you alert
Another problem is you try to return result from alert function which does not provide result

Related

I keep getting "undefined" as an error, function not returning anything

I've been trying to figure out why I keep getting undefined, and I don't really see why, I know that I would get undefined if my function isn't returning anything, but in this case, even with returning in two places, I am still getting undefined. I think that the reason for that is that maybe I am not returning the value from innerFunc properly? I'm thinking maybe there's something wrong with my syntax here: innerFunc(arg) but I'm not sure how it is wrong, or what to change it to.
Any help would be appreciated.
Instructions:
Create a function "fastCache" that takes one argument (a function) and returns a function. When fastCache is invoked it creates an object that tracks calls to the returned function, where each input to the returned function is associated with its output. Every subsequent call to that returned function with the same argument will return the output directly from the object, instead of invoking the original function again.
function fastCache(func) {
const obj = {};
function innerFunc(arg) {
for (const key in obj) {
if (key === arg) {
return obj[arg]
} else {
obj[arg] = innerFunc(arg)
return innerFunc(arg)
console.log(obj[arg])
console.log(arg)
console.log(innerFunc(arg))
}
}
}
return innerFunc
// console.log(innerFunc(arg))
}
//example:
//SINGLE ARGUMENT CASE
const multiplyBy2 = num => num * 2;
const cachedMultiplyBy2 = fastCache(multiplyBy2);
If you want a simple cache function:
function cacheWrapper(fn) {
let cache = { };
return arg => {
if (cache.hasOwnProperty(arg)) {
return cache[arg];
}
return cache[arg] = fn(arg);
};
}
function test(x) {
return x * 2;
}
let testCached = cacheWrapper(test);
console.log(testCached(2));
console.log(testCached(2));
The key here is hasOwnProperty which will tell you if that cache slot is occupied or not. This avoids having to fumble around and find out the hard way with iteration.
If the number isn't already in the object, it won't be found in the loop.
function fastCache(func) {
const obj = {};
function innerFunc(arg) {
for (const key in obj) {
if (key === arg) {
return obj[arg]
}
}
obj[arg] = func(arg)
return func(arg)
}
return innerFunc
// console.log(innerFunc(arg))
}
//example:
//SINGLE ARGUMENT CASE
const multiplyBy2 = num => num * 2;
const cachedMultiplyBy2 = fastCache(multiplyBy2);
const result = cachedMultiplyBy2(8);

JavaScript Functions , return undefined

Hello Everyone hope you all doing great ,
this is my code , function with name and callback
taking the name to callback function to make return the name and console.log it
if i
function doSomething(name,callback) {
callback(name);
}
function foo(n) {
return n;
}
var val = doSomething("TEST",foo);
console.log(val);
i got undefined .
if i
function doSomething(name,callback) {
callback(name);
}
function foo(n) {
console.log(n);
}
var val = doSomething("TEST",foo);
that works fine and console.log the TEST name .
So why the return not working ?
thanks
Your doSomething() function doesn't return anything, which means an assignment using it will be undefined. But, that's not really the problem here.
The underlying problem is that you seem to be mixing two different data processing patterns here: if you're writing purely synchronous code, then use returning functions (which immediately return some value). If you need asynchronous code, then use a callback (which will "eventually" do something). Mixing those two patterns is a recipe for problems and frustration:
Either:
don't name your function a "callback", and have it return its processed value, or
make the callback responsible for doing whatever it is you were going to do with val.
Case 1:
function doSomething(data, processor) {
return processor(data);
}
function passThrough(v) { return v; }
var val = doSomething("test", passThrough);
// immediately use "val" here in for whatever thing you need to do.
Case 2:
function doSomething(data, callback) {
// _eventually_ a callback happens - for instance, this
// function pulls some data from a database, which is one
// of those inherently asynchronous tasks. Let's fake that
// with a timeout for demonstration purposes:
setTimemout(() => callback(data), 500);
}
function handleData(val) {
// use "val" here in for whatever thing you need to do. Eventually.
}
doSomething("test", handleData);
And if you want to go with case 2, you really want to have a look at "Promises" and async/await in modern Javascript, which are highly improved approaches based on the idea of "calling back once there is something to call back about".
2021 edit: a third option since original writing this answer is to use the async/await pattern, which is syntactic sugar around Promises.
Case 3:
async function doSomething(input) {
// we're still _eventually_ returning something,
// but we're now exploiting `async` to wrap a promise,
// which lets us write normal-looking code, even if what
// we're really doing is returning a Promise object,
// with the "await" keyword auto-unpacking that for us.
return someModernAsyncAPI.getThing(input);
}
function handleData(val) {
// ...
}
async function run() {
const data = await doSomething("test");
handleData(data);
}
run();
function doSomething(name,callback) {
callback(name);
}
function foo(n) {
console.log(n);
return n;
}
var val = doSomething("TEST",foo);
Take a look at above code. When you call doSomething, which internally executes foo it prints on the console because thats what console.log is for. However, after this statement it returns n as well which then is received in doSomething. But its not being returned. To put it simply, what you are mainly doing is
function doSomething(name,callback) {
const returnValue = callback(name);
}
If you call the above method, it will return undefined. To make it return correct value, you have to call "return returnValue". Similary you have to say
return callback(name)
Hope this helps.
Happy Learning
Inorder to assign the returning value/object of a function(in this case doSomething, it should have a return statement. Else the function returns nothing, so when you assign that to val, it is still undefined.
So you modify your code like this:
function doSomething(name,callback) {
return callback(name);
}
function foo(n) {
return n;
}
var val = doSomething("TEST",foo);
console.log(val);
undefined is implicitly returned if you don't have a return in your function.
when you call var val = doSomething("TEST",foo), you are aligning the return value of doSomething to val, which is undefined.
function doSomething(name,callback) {
return callback(name);
}
function foo(n) {
return n;
}
var val = doSomething("TEST",foo);
console.log(val);

About javascript function valueOf\toString and 'curry' function behaved differently in Chrome,Firefox and node enviroment

I wrote a function:
function add(){
let arr = [];
arr = arr.concat(Array.prototype.slice.apply(arguments))
let fun = function(){
arr = arr.concat(Array.prototype.slice.apply(arguments))
return fun
}
fun.toString = function(){
console.log(222)
return arr.reduce(function(total, num){
return total+num
}, 0)
}
return fun
}
console.log(add(1,2)(2,3)(3))
This is in Chrome:
enter image description here
Two questions:
In first line, why is 'f 11' ,not '11'?
Why output 'f 11' firstly, not '222', I think the type conversion should execute firstly, and then output computed result on console.
Another strange thing, it is the result in Firefox with same codes:
enter image description here
And the result in node environment:
enter image description here
I do not understand why, it seems in FF and node, has not executed the computed operation.
Please help me...Thanks so much!
At first you can beautify the whole code a bit:
function add(..arr){
function fun(...args){
arr.push(...args);
return fun
}
fun.toString = function(){
return arr.reduce((total, num) => total + num)
};
return fun;
}
And as you noticed correctly, logging a function is completely up to the environment. Firefox and Node return the code of the function, while Chrome does sth like:
out( "f" + add.toString())
so our toString function gets called and something is logged. To have a consistent behaviour between the different environments we could call toString explicitly:
console.log(add(1)(2)(3).toString());
This can be inferred:
console.log("" + add(1)(2));
If what you want is add that is both variadic and curried (which I still think is weird), just do this:
const add = (...args) => {
let accum = args;
let f = (...fargs) => {
if (!fargs.length) {
return accum.reduce((a, b) => { return a + b; }, 0);
} else {
accum = accum.concat(fargs);
return f;
}
};
return f;
};
add(1,2,3)(); // 6
add(1)(2,3)(); // 6
add()(); // 0
Now you just call the returned function with no arguments to get the value out. You could toy with it to make it more performant (e.g. by using .push in a loop instead of .concat) but this should work.

How to access a function within another function on a string?

In javascript, how to access function "look" in this sample.
This is the function:
var func = "(function() {function look() {return 12;};return 'OK'})()";
I try this
eval(func);
And return "OK", correct. But I need the return of the "look" funtion (return 12.
Is this possible?
Yes. Change return "OK" to return look().
var func = "(function() { function look() { return 12; } return look(); })()";
eval(func); // 12
Otherwise your look() function is never executed and, even if it currently was, you're returning the string "OK" anyway.
I'm not really sure why you're wanting to use eval() though. Why don't you simply:
var func = function() { function look() { return 12; } return look(); }
func(); // 12
Yes!,
You can, Using closure you can achieve your answer.
var func="(function () { return function look() { return 12; } })()";
var cFun=eval(func);
cFun(); //12

What is the point of the return in Javascript?

I've recently been learning javascript and came across the return statement or whatever it is. I've watched tutorials on it and I still don't get the point of it. Do I ever need it for anything and can someone please explain in detail what it does? Thanks
The return statement has multiple uses:
1) Return early from your function before the end of the function body. This is often within a branch of the code as in:
function whatever()
// do some things
if (err) {
console.log(err);
return err;
}
// other code here that will not execute if there was an err
}
2) Return a specific value back to the caller as in:
function add(a, b) {
return a + b;
}
var sum = add(3,4);
console.log(sum); // will show 7
The return statement in javascript can be used by itself to just exit the current function call and not return a specific value or it can be used to return a specific value.
If no return statement is present in a function, then the function will execute to the end of the function body and will return automatically at the end of the function body.
The only point of it is to send a value (string, number, or other values) from your function so that is can be used outside the function. Instead you can use global variables but that takes up more code, so it's a shortcut instead.
jfreind00 above said that you can use it to branch off early but that's what break. Here is the syntax:
function HiThere(something) {
if (something === true) {
break;
}
alert("Hi");
}
In this case, if something is true, then exit the function, if not then say hi.
Return would be like this:
function HiThere(something) {
if (something === true) {
return "Value to return: " + something;
}
alert("Hi");
}
In this case, if something is true, then exit the function and return Value to return: true, if not then say hi but don't return anything back.
Here's the most common use of the return statement:
document.getElementById("demo").innerHTML = myFunction(5,3); // Call a function
// to set the "demo" element to the value of: 5 * 3
function myFunction(a,b) {
return a * b;
}
<p id="demo"></p>
Use it to return a value. Your value could be "Yes", "false", "hello world" or "eat dirt." You can then run certain code based on the returned value.
Hope this helps!

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