Return Value Using Arrow Syntax

So to my understanding which is obviously wrong at this moment in time is that, return arg => arg*2 is the same as return (arg)=>{arg*2} I always assumed arrow functions a

Solution 1:

arrow function works differently here:-

(x)=> x*2 ; // dont have to return anything, x*2 will be returnedisnot same as 
(x) =>{x*2}
//here you need to return something otherwise undefined will be returned

Solution 2:

When you use {} you must set return

return(arg)=>{return arg*2}

Solution 3:

The arrow function only supplies the return automatically if you have an expression after the arrow. If the arrow is followed by a curly brace, it's treated as the braces around a function body, so you have to write return explicitly, i.e.

arg => arg * 2

is equivalent to:

(arg) => { returnarg * 2; }

functionaddTwoDigits(firstDigit) {
  return(secondDigit) => {
    return firstDigit + secondDigit
  }
}
let closure = addTwoDigits(5);
console.log(closure(5))

Solution 4:

You need a return statement if the body of your arrow function is wrapped in { ... }.

Since yours is a single expression, you can skip the {-} and return. But if you have the {-}, you need the return statement.

The parentheses are not an issue here. You need them if you have more than a single parameter. With one, they're optional.