Forum

Posted: **10 Nov 2020, 08:09**

Why the function.call() behaves differently with and without 'this', in JavaScript?
The result with 'this' in test.call() is same when 'this' is replaced by 'undefined'.

Code: Select all

function test(a,b){
  console.log(a+b);
}

let args = [1,2]
test.call(this, ...args); // Output: 3
test.call(undefined, ...args); // Output: 3

But without 'this' in test.call(), the result is NaN.

Code: Select all

function test(a,b){
  console.log(a+b);
}

let args = [1,2]
test.call(...args); // Output: NaN

Posted: **10 Nov 2020, 08:59**

The call() function require the first parameter as 'this' object, if you do not need it, just pass null.

Code: Select all

test.call(null, ...args);

In your code, you pass args into the 'this' in test function not into (a, b).
You can add console.log(this) to checkout the different between two cases.

Code: Select all

function test(a,b){
  console.log(this) // number 1
  console.log(a) // number 2
  console.log(b) // undefined
  console.log(a+b) // 2 + undefined will be NaN
}

let args = [1,2]
test.call(...args);

So, that means the first argument will be 'this'.
If you pass the string in first argument, your 'this' will be that string.

Code: Select all

function test(a,b){
    console.log(this) // "marplo"
    console.log(a) // number 1
    console.log(b) // number 2
}

let args = [1,2]
test.call('marplo', ...args);

Forum

Problem with 'this' in function.call() in JavaScript

Problem with 'this' in function.call() in JavaScript

Problem with 'this' in function.call() in JavaScript