Swapping of two variables using function & without. function in JS but without creating third variable.

As we all know to swap two variables we choose to create third variable and then we swap their values like let me show you with example :
var x = 1;
var y = 2;
temp = x;
and then we go further , now here we'll be showing you different methods to swap variables in javascript without making use of third variable.

function swap(value) {
    value.first = value.first + value.second;
    value.second = value.first - value.second;
    value.first = value.first - value.second; 
}
// declared an object named value which has two keys - first and second corresponding to the first and the second values respectively

var value = {
    first: 1,
    second: 2
}

console.log(value.first, value.second) //Prints 1 2
swap(value);
console.log(value.first, value.second); //Prints 2 1

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Second Way by creating function again

function swap(x, y) {
  return [y, x]
}
var x=1, y=2;
console.log(x+" "+y);

[x, y] = swap(x, y);
console.log(x+" "+y)

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Both ways you get the same result

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Without Creating function

var x = 1, y = 2;
console.log(x , y)
x = x + y;
y = x - y;
x = x - y;
console.log (x+" "+y); // 2 1

Array destructuring method

var x=1; y=2;
console.log(x,"",y);
[x ,y] = [y ,x];
console.log(x,"",y)

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This method will also result you the same result

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If you want to return in array format

var x=1; y=2;
console.log(x,"",y) // 1 2
function swap()
{
  var t = x;
  x = y;
  y = t;
  return [x , y]
} 
console.log(swap(x+"  "+y)); //[2 1]

Using bitwise xor operator:

let x = 1, y = 2;
x = x ^ y;
y = x ^ y;
x = x ^ y;

console.log(x ,y);