Correction for the last question - At 22:15 args.length is 1, which of course it is, then I said length of function is 4, and highlighted the four curried functions at line number 22. Which is kind of not the case here, Here, func is that sum function I passed as prameter to original curry function and func.length would be length or number of parameters that func (sum function) expects. Which in this case sum function expects four formal parameters which are: a, b, c, d. So four comes from here, not from the curried functions in line 22. Sorry for the explanation gap.
it was the length of the first argument as callback function on the first call. if the arg.length == curryArg.length ? then only call the callback function.. this one is actually very usefull.
this is one of the coolest thing I learn !! I have all those chained http post calls, now I can do post(data1)(data2)(data3)((res) => { something}) without using promise and async/await and at the same time, no callback hell. I can throw error and use try/catch like async/await does.
Currying is a function that takes one argument at a time and returns a new function expecting the next argument. It is a conversion of function from callable as f(a,b) to f(a)(b).
you should also explain this The length data property of a Function instance indicates the number of parameters expected by the function. for some folks it could be possible they couldnt understand how func.length is coming
Thank you so much for educating and your videos were super mind game. I know you are intended to communicate in simple English and react language. Since I am a beginner in JS, just excited to share this the one n only question I solved better. const arithOp = { add: "+", subtract: "-", divide:"/", multiply:"*" } function one(type){ return function(lhs){ return function(rhs){ const resultString = `lhs ${arithOp[type]} rhs`; console.log(eval(resultString)); } } } one("add")(3)(3);
Great video, but i guess it was slightly confusing when you said, func.length. At 22:15 you said args.length is 1, which of course it is, then you said length of function is 4, and highlighted the four curried functions at line number 22. Which is kind of not the case here, Actually here func is that sum function you passed as prameter to original curry function and func.length would be length or number of parameters that func (sum function) expects. Which in this case sum function expects four formal parameters which are: a, b, c, d. So four comes from here, not from the curried functions in line 22. As according to MDN, length is a property of a function object, and indicates how many arguments the function expects, i.e. the number of formal parameters. This number excludes the rest parameter and only includes parameters before the first one with a default value. Nonetheless great video!
The best thing about this series is that "You are covering questions topic wise in each separate video", So we can say we covered few question on this or that topic.
i believe that your example for the "curry" function allows for passing multiple arguments in each subsequent invocation of a curried function. Isn't it more accurate to allow only one argument in each curried call? by removing the rest and spread operators on "next", on lines 11 + 12. 24:16
Nice effort earlier I though it used to be a cuury made inside the kitchen only but in js how to used with the function its great ,brother you are great
Loved the video bro, although i think there is no need to do: 1. the rest and spread of next arguement 2. greater condition check i.e. (args.length >= func.length) OR correct me, if i m wrong somewhere function curry(func) { return function curriedFunc(...args) { if (func.length === args.length) { return func(...args); } else { return function (next){ return curriedFunc(...args, next); } } } }; Also for those who don't know - function.length -> returns the EXPECTED number of arguments a function is expecting excluding the default and rest parameter arguments.length -> returns the ACTUAL number of arguments passed to a function
This time in Cars24 i was asked this question. You need to curry add if n number of arguments are placed in any manner after the function: sum(1)(2)()()()(3)
Only if the last argument is empty we can go for Infinite currying as the tutorial explained if the middle argument is empty it stops there itself and gives result. Anyway what was the answer you gave and how can we solve this problem?
Thank you so much sir, it is very much helpful and all your videos have helped to understand the javascript very well especially all your interview videos sir, thank you so much sir
Hey piyush there is catch in function add(a){ return function(b){ if(b) return add(a+b) return a; } } example console.log(add(10)(0)()); if b = 0 then it does not return any function so its throw an error called TypeError: add(...)(...) is not a function here is a proper solution function add(a){ return function(b){ if(b !== undefined) return add(a+b) return a; } } thank you _/\_
When the interviewer asked us to write a function for multipler6 (if I have given 3 it should return 18, if 6 then 36 ) this is called partial application. const mul = (a=6) => { return (b) => { return a * b; }; }; const mul6 = mul(); console.log(mul6(3)); console.log(mul6(13)); console.log(mul6(9));
in question 5- Manipulating Dom we can get the ref of element in one variable and then keep on updating the content, so why curring adding more sense here?
Interview has asked me so its become more difficult to write function if we use currying so why we are using it like : function add(a,b,c){ return a+b+c; } And same if we do currying it will take 2 internal function and than it will return us value. So bhaiya how should we answer this questions??
const sum = function(a) { return function(b) { if (b) { return sum(a + b); } else { return a; } } } yes we can write that in one line: const add = a => b => b ? add(a + b) : a;
My Interview asked me this question -> Write a function sum that takes n arguments and can be called n times. Note - Function can take n arguments and can be called n times. For example - sum(1,3,5,1)(1,2,3,4,5)(1)(1,2)(1,1,1,1,1,1)(1,4,5)(12)(11)(50, 51, 52) console.log(sum.....) => 221 It was for SDE - 1 Role. I still not able to understand this problem.
function sum(...args){ let a = args.reduce((a,b)=>a+b,0); return function (...args){ let b = args.reduce((a,b)=>a+b,0); if(b) return sum(a+b); return a; } }
So we will need to pass () as blank to stop the recursion in 13:31 otherwise the output is always a function . Is there any way we can do this without adding empty () in the end??
Hi, you don't need to pass () as blank. What will stop the recursion is the "if" statement once it will be true, meaning once the number of args will match the number of functions.
@@dkatorzaify If u will quickly run this in browser console, you can see it is always returning a function. Logic seems okay but its not working, without the ()
@@akshitatyagi5234 This is strange... I copied this to the console and it does work. function curry(func) { return function curriedFunc(...args) { if(args.length >= func.length) { return func(...args) } else { return function(...next) { return curriedFunc(...args,...next); } } } } const sum = (a,b,c,d) => a+b+c+d const totalSum = curry(sum) console.log(totalSum(1)(2)(3)(4))
Hi @RoadsideCoder , with infinite currying code, its returning only Function ..not the addition, Could you please clarify. function add(a){ return function(b){ if(b) return add(a + b); return a; } } console.log(add(4)(5)(7)); OUTPUT: [Function]
it is returning function because you should call like this console.log(add(4)(5)(7)()) ; without (), function does not get invoked and you get only function body
This will give you an error when you pass b=0 to handle all cases use below code function add(a) { return function (b) { return b >= 0 ? add(a + b) : a; }; } console.log(add(1)(2)(4)(0)());
Hi, I've written below code. Is it write wrt to currying?' const evalute = a => { switch (a) { case 'sum': return function (b) { return function (c) { return b + c; }; }; case 'sub': return function (b) { return function (c) { return b - c; }; }; default: break; } }; console.log(evalute('sum')(5)(3)); console.log(evalute('sub')(5)(3));
Correction for the last question -
At 22:15 args.length is 1, which of course it is, then I said length of function is 4, and highlighted the four curried functions at line number 22.
Which is kind of not the case here,
Here, func is that sum function I passed as prameter to original curry function and func.length would be length or number of parameters that func (sum function) expects.
Which in this case sum function expects four formal parameters which are:
a, b, c, d.
So four comes from here, not from the curried functions in line 22.
Sorry for the explanation gap.
Exactly what I was going to comment and ask! Glad to find your comment 😅
it was the length of the first argument as callback function on the first call.
if the arg.length == curryArg.length ? then only call the callback function..
this one is actually very usefull.
You are putting so much effort to make such topics easy for us. Thanks for putting content like this. We really appreciate it.
❤️❤️ Thank you so much, greatful
this is one of the coolest thing I learn !! I have all those chained http post calls, now I can do post(data1)(data2)(data3)((res) => { something}) without using promise and async/await and at the same time, no callback hell. I can throw error and use try/catch like async/await does.
Currying is a function that takes one argument at a time and returns a new function expecting the next argument. It is a conversion of function from callable as f(a,b) to f(a)(b).
you should also explain this The length data property of a Function instance indicates the number of parameters expected by the function. for some folks it could be possible they couldnt understand how func.length is coming
yes... by the way thank you for explaining
Thank you so much for educating and your videos were super mind game.
I know you are intended to communicate in simple English and react language.
Since I am a beginner in JS, just excited to share this the one n only question I solved better.
const arithOp = {
add: "+",
subtract: "-",
divide:"/",
multiply:"*"
}
function one(type){
return function(lhs){
return function(rhs){
const resultString = `lhs ${arithOp[type]} rhs`;
console.log(eval(resultString));
}
}
}
one("add")(3)(3);
Awesome! Thanks for adding this approach here
Great next make on OOPS concept and prototype related
Great video, but i guess it was slightly confusing when you said, func.length.
At 22:15 you said args.length is 1, which of course it is, then you said length of function is 4, and highlighted the four curried functions at line number 22.
Which is kind of not the case here,
Actually here func is that sum function you passed as prameter to original curry function and func.length would be length or number of parameters that func (sum function) expects.
Which in this case sum function expects four formal parameters which are:
a, b, c, d.
So four comes from here, not from the curried functions in line 22.
As according to MDN,
length is a property of a function object, and indicates how many arguments the function expects, i.e. the number of formal parameters. This number excludes the rest parameter and only includes parameters before the first one with a default value.
Nonetheless great video!
Yes, my bad.. That was a little gap in my explanation, I'll add a pinned comment for correcting this part.
@@RoadsideCoder Cool!
Brother, you are a gem. Thank you so much.
Now , watching this type of video, lets hit the interview.. You are best of best .. ,🔥
All the best 😎🔥
Thank you sir, your video's is really helpful please continue this interview series
Yes, More on the way!
The best thing about this series is that "You are covering questions topic wise in each separate video", So we can say we covered few question on this or that topic.
Glad it helped you!
i believe that your example for the "curry" function allows for passing multiple arguments in each subsequent invocation of a curried function. Isn't it more accurate to allow only one argument in each curried call? by removing the rest and spread operators on "next", on lines 11 + 12. 24:16
🔴 Get my Complete Frontend Interview Prep course - roadsidecoder.com/course-details
Nice effort earlier I though it used to be a cuury made inside the kitchen only but in js how to used with the function its great ,brother you are great
Deserve a million subscribe
Loved the video bro, although i think there is no need to do:
1. the rest and spread of next arguement
2. greater condition check i.e. (args.length >= func.length)
OR correct me, if i m wrong somewhere
function curry(func) {
return function curriedFunc(...args) {
if (func.length === args.length) {
return func(...args);
} else {
return function (next){
return curriedFunc(...args, next);
}
}
}
};
Also for those who don't know -
function.length -> returns the EXPECTED number of arguments a function is expecting excluding the default and rest parameter
arguments.length -> returns the ACTUAL number of arguments passed to a function
thanku for function.length explaination
Good Point 👍👍 But if i do not spread the next argument, the following function will not work --> curriedSum(1)(2,3);
Hope you understand.
This is gold ... 👍👍 Such a quality content . .. keep going brother...
🙏🙏
My answer for currying
function sum(args){
if(args == undefined) {
let currentSum = sum.currentSum;
sum.currentSum = 0;
return currentSum;
}
sum.currentSum = (sum.currentSum || 0) + args;
return sum;
}
Thanks for u r video with clear explanation.
impressive teaching methodology..
This time in Cars24 i was asked this question. You need to curry add if n number of arguments are placed in any manner after the function:
sum(1)(2)()()()(3)
Only if the last argument is empty we can go for Infinite currying as the tutorial explained if the middle argument is empty it stops there itself and gives result. Anyway what was the answer you gave and how can we solve this problem?
@@SIVAREDDY-e4l Yes using recursion, you have to check whether the arguments is equal to the function length.
Thanks for teaching a new concept
Thank you so much sir, it is very much helpful and all your videos have helped to understand the javascript very well especially all your interview videos sir, thank you so much sir
Great Work bro, keep it up
Thanks 🔥
Great Quality Content....
Hey piyush there is catch in
function add(a){
return function(b){
if(b) return add(a+b)
return a;
}
}
example
console.log(add(10)(0)());
if b = 0 then it does not return any function so its throw an error called
TypeError: add(...)(...) is not a function
here is a proper solution
function add(a){
return function(b){
if(b !== undefined) return add(a+b)
return a;
}
}
thank you _/\_
Great buddy
@@MrCoder-u9y thanks
When the interviewer asked us to write a function for multipler6 (if I have given 3 it should return 18, if 6 then 36 ) this is called partial application.
const mul = (a=6) => {
return (b) => {
return a * b;
};
};
const mul6 = mul();
console.log(mul6(3));
console.log(mul6(13));
console.log(mul6(9));
in question 5- Manipulating Dom we can get the ref of element in one variable and then keep on updating the content, so why curring adding more sense here?
Thanks you for explaining it very well, I really understand it now 😅...
Super useful video. Thanks
The last question was epic 🤟
Bhaiya please reply in timestamp 15:58 actually I think the example you provide is actually behave like a closure not currying tell me if i am wrong
Really appriciate you content
Isnt question 5 (dom manipulation) here an example of using closures instead of currying? How does currying link to this question?
Interview has asked me so its become more difficult to write function if we use currying so why we are using it like :
function add(a,b,c){
return a+b+c;
}
And same if we do currying it will take 2 internal function and than it will return us value.
So bhaiya how should we answer this questions??
Please make a video on throttling and debouncing in JavaScript 🥺🙏
Yes that's in the way!
Love your videos. 💙 Please bring fast too
Thank you!
07:48, you can use Switch case. It's little better than nested If else loops.
Great video and very well explained
Thanks ❤️
const sum = function(a) {
return function(b) {
if (b) {
return sum(a + b);
} else {
return a;
}
}
}
yes we can write that in one line:
const add = a => b => b ? add(a + b) : a;
My Interview asked me this question ->
Write a function sum that takes n arguments and can be called n times.
Note - Function can take n arguments and can be called n times.
For example -
sum(1,3,5,1)(1,2,3,4,5)(1)(1,2)(1,1,1,1,1,1)(1,4,5)(12)(11)(50, 51, 52)
console.log(sum.....) => 221
It was for SDE - 1 Role.
I still not able to understand this problem.
const infiniteSum = (...a) => {
return (...b) => {
const currSum = [...a, ...b].reduce((sum, val) => sum + val, 0);
return b.length ? infiniteSum(currSum) : currSum;
};
};
@@Abhishek_Sawant Hey thanks man :)
@@Abhishekverma-yj1or 👍😅
function sum(...args){
let a = args.reduce((a,b)=>a+b,0);
return function (...args){
let b = args.reduce((a,b)=>a+b,0);
if(b) return sum(a+b);
return a;
}
}
in my first call the Hr asked me waht web application do you use....i was noy aware what web applications are...? could you explain me
So we will need to pass () as blank to stop the recursion in 13:31 otherwise the output is always a function . Is there any way we can do this without adding empty () in the end??
Hi, you don't need to pass () as blank. What will stop the recursion is the "if" statement once it will be true, meaning once the number of args will match the number of functions.
@@dkatorzaify If u will quickly run this in browser console, you can see it is always returning a function. Logic seems okay but its not working, without the ()
@@akshitatyagi5234 This is strange... I copied this to the console and it does work.
function curry(func) {
return function curriedFunc(...args) {
if(args.length >= func.length) {
return func(...args)
} else {
return function(...next) {
return curriedFunc(...args,...next);
}
}
}
}
const sum = (a,b,c,d) => a+b+c+d
const totalSum = curry(sum)
console.log(totalSum(1)(2)(3)(4))
@@dkatorzaify I am talking about the example at 13:31
sir i have a doubt in console.log(add(2)(3)(4)); when remove the last parenthesis it showing output as function [anonymous] , sir what will be reason?
Thank you
Subscribed❤
Bhai Make video on spread operator tricks
Hi @RoadsideCoder , with infinite currying code, its returning only Function ..not the addition, Could you please clarify.
function add(a){
return function(b){
if(b) return add(a + b);
return a;
}
}
console.log(add(4)(5)(7));
OUTPUT: [Function]
it is returning function because you should call like this console.log(add(4)(5)(7)()) ; without (), function does not get invoked and you get only function body
How much can we demand as front end developer with 1 year experience?
Minimum 15 LPA
In manipulating DOM example, why don't we simply create a fucntion instead of returning a fucntion
Can u please tell how to solve a error while making a project
how much can we demand for frontend developer role with 3.9 yrs of experience?
Depends on your skills, sky is the limit.
@@RoadsideCoder I have experience and knowledge in JS, ReactJS and React Native
This will give you an error when you pass b=0
to handle all cases use below code
function add(a) {
return function (b) {
return b >= 0 ? add(a + b) : a;
};
}
console.log(add(1)(2)(4)(0)());
Now i am confused between Closure and Currying 😅
Hi,
I've written below code. Is it write wrt to currying?'
const evalute = a => {
switch (a) {
case 'sum':
return function (b) {
return function (c) {
return b + c;
};
};
case 'sub':
return function (b) {
return function (c) {
return b - c;
};
};
default:
break;
}
};
console.log(evalute('sum')(5)(3));
console.log(evalute('sub')(5)(3));
Follow dry(Don't Repeat Yourself)
done ... !
Bro jaldi jaldi video daalo
how are you sir please make a project on MERN as like dosti chat app
Sure :)
@@RoadsideCoder thanks love you sir
Please put the on github sir .
🔥🔥🔥
*okay cool*
I want theory related and code based questions with pdf
Sure, I'll create a github repository for it!
now legends will only subscribe you because of "69" years of goodluck 🤣
Are u that legend?
@@RoadsideCoder yes I am😂 and thank you from the depth of my heart. for creating such awesome content 💙
@@jaisharma545 welcome 🙏
0:31 🤣
Why 69 years of goodluck..... means why 69 only 😁😁😁