Kevin Powell - Community•3y ago

I need to understand this part of the code

let x = "hello my name is chrono" const finalSentence = x.replace(/\S\w | \s\w/g, letter => letter.toUpperCase()); >>>>>>>>>>>>> I need to understand this part of the code >>>>>>>>>>>> letter => letter.toUpperCase() Is it like writing this function letter(letter) { letter.toUpperCase() return letter }

12 Replies

13eck•3y ago

Exactly, yes It's called an arrow function. Give me a sec to find the MDN link

13eck•3y ago

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/Arrow_functions

Arrow function expressions - JavaScript | MDN

An arrow function expression is a compact alternative to a traditional function expression, but is limited and can't be used in all situations.

chrono1913OP•3y ago

but I want to know for that part of the code, could you write an outside function and pass it into that part of the code like const finalSentence = x.replace(/(^\w{1})|(\s+\w{1})/g, letter()); function letter(letter) { letter.toUpperCase() return letter }

Joao•3y ago

Yes, you can do that as well But, instead of invoking the function like in your question, you just provide the reference to that function:

// before
const finalSentence = x.replace(/\S\w | \s\w/g, letter());

// after
const finalSentence = x.replace(/\S\w | \s\w/g, letter);

// before
const finalSentence = x.replace(/\S\w | \s\w/g, letter());

// after
const finalSentence = x.replace(/\S\w | \s\w/g, letter);

chrono1913OP•3y ago

but its still giving me an error did I write the function correctly

Joao•3y ago

and the error is? always read the errors, we can't really do anything without this information

chrono1913OP•3y ago

o forgot the closing bracket but not the output is “helundefinedundefinednaundefinedundefinedchrono”

Joao•3y ago

And what do you think may cause that What is the expected output vs the actual output?

chrono1913OP•3y ago

i think its the function there isn’t any parameter in the function right> ? "Hello My Name Is Chrono"

Joao•3y ago

The issue is not the function, it's the pattern. Read the description on how replace works when you provide a function and you will see the problem: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace#specifying_a_function_as_the_replacement Hint: in the output you are seeing a bunch of "undefined" in between the text

chrono1913OP•3y ago

when you say pattern do you mean the REGEXP /(^\w{1})|(\s+\w{1})/g thats what I was using

MarkBoots•3y ago

const input = "hello my name is chrono";
const output = input.replace(/\b\w/g,(m)=>m.toUpperCase());
console.log(output) // "Hello My Name Is Chrono"

const input = "hello my name is chrono";
const output = input.replace(/\b\w/g,(m)=>m.toUpperCase());
console.log(output) // "Hello My Name Is Chrono"

\b word boundary \w any word character /g global, all matched (m)=> ... for each match So every first word character after a word boundary will be matched and replaced or if you want the replacement outside the replace function

const input = "hello my name is chrono";
const output = input.replace(/\b\w/g, replaceFunction);
console.log(output) // "Hello My Name Is Chrono"

function replaceFunction(match){
  return match.toUpperCase()
}

const input = "hello my name is chrono";
const output = input.replace(/\b\w/g, replaceFunction);
console.log(output) // "Hello My Name Is Chrono"

function replaceFunction(match){
  return match.toUpperCase()
}

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace

String.prototype.replace() - JavaScript | MDN

The replace() method returns a new string with one, some, or all matches of a pattern replaced by a replacement. The pattern can be a string or a RegExp, and the replacement can be a string or a function called for each match. If pattern is a string, only the first occurrence will be replaced. The original string is left unchanged.

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I need to understand this part of the code