CODEWITHSUNDEEP
javascript
Sugumar Venkatesan
Sugumar Venkatesan

Reputation: 4028

How the call function works with slice function?

I don't understand the call function in javascript

function multiply(){
    console.log(arguments);
    console.log([].slice.call(arguments));
}

console.log(multiply(2,3));

The above function prints

// [Arguments] { '0': 2, '1': 3 }
// [ 2, 3 ]

So I see the value passed to call function is an following object

//[Arguments] { '0': 2, '1': 3 }

So I tried the following but it prints empty array,

function multiple(){
    console.log(arguments);
    console.log([].slice.call({ '0': 2, '1': 3 }));
}

console.log(multiple(2,3));

// [Arguments] { '0': 2, '1': 3 }
// []

Can anybody explain me why the call works in first case, and why it is not working in second case?

How exactly call and slice work?

Upvotes: 0

Views: 88

Answers (1)

jason_r
jason_r

Reputation: 345

In the first case you're simply passing the two numbers into call and there isn't an object being referenced as the first argument to call, so it will then identify those numbers as arguments, which is represented as an array-like object (it has a length) when logged. In the second case you're passing in an actual object (yes, identical looking to what ends up getting logged in the first case) as the first argument, but since it's an object, call thinks that object is what you're referencing as the this value that would be used for any subsequent argument values being passed in, of which you have none so it just returns an empty array. I'm relatively newer to JavaScript but from what I've seen the form [].slice.call(x, y, z) has often been used in the past to convert array-like lists into actual arrays, but these days with ES6 JavaScript you can just use a spread operator or Array.from() to do this much more easily. This post has more info on the slice and call type of pattern.

Upvotes: 1

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