CODEWITHSUNDEEP
javascript
김진석
김진석

Reputation: 19

about javascript arrow function expression

I am Korean, so I hope you could understand it with my awkward English skills

I am studying arrow function expressions

and here is my question with source code

var arguments = [1, 2, 3];
var arr = () => arguments[0];

console.log(
  arr()
)


function foo(n){
  var f = () => arguments[0] + n
  console.log(arguments[0]) //5
  console.log("----arguments[0]")
  return f();
}
console.log(
  foo(5)
)

console.log("---------------------")

function fooo(n){
  var f = (...arguments) => arguments[0] + n
  console.log(...arguments) //5
  console.log("----...arguments")
  return f(2);
}

console.log(
  fooo(5)
)

I don't get why the second function's console.log = 10 and third function's = 7 can anyone explain to me the order code process and why that output is? thank you.

Upvotes: 0

Views: 78

Answers (2)

Charlie
Charlie

Reputation: 23778

In JS, every conventional function has a built-in object called arguments. However, Arrow functions do not have this built-in object. So, if referred to arguments from within an Arrow function, the reference automatically goes to any external variable declared by that name.

The result you see is due to this effect. In regular function calls. your reference to the arguments global variable actually refers to their in-built arguments object.

The following code demonstrates this effect.

let arguments = 'this is a string';

function argumentsTest() {
   console.log(arguments);
}


let argumentsArrowTest = () => {
   console.log(arguments);
}


argumentsTest(10);        //{0: 10}
argumentsArrowTest(10);   // this is a string

Upvotes: 2

CertainPerformance
CertainPerformance

Reputation: 370699

In an arrow function, arguments, like this, will refer to either

  • an outer identifier named arguments (though it's very unusual to call a variable arguments due to name collision), or
  • the arguments provided to a full-fledged function

whichever is closer, lexically.

In foo,

var f = () => arguments[0] + n

is an arrow function, so arguments refers to an outer binding. The outer environment is

function foo(n){
  var f = () => arguments[0] + n

so arguments refers to the arguments provided to foo. Here, it's similar to doing

function foo(n){
  const theArguments = [n];
  var f = () => theArguments[0] + n

And since the argument passed to foo is 5, that plus n (the same argument) is 10.

In the third function, the same logic applies. Inside an arrow function, arguments refers to the closest identifier named arguments, or the closest function. Here, it's the parameters:

var f = (...arguments) => arguments[0] + n

Since f is called with a single argument, this simplifies to

var f = (arg) => arg + n

That argument is 2, and the n is what's passed to fooo, which is 5. Add them together, you get 7.

Upvotes: 0

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