function add(x=1, y=2){
return x + y;
}
console.log(add(y=4));
Output I am getting is 6, Why? How does this work?
I was expecting 5 as y=4 and x will retain its default value.
Is there any other method of doing this without using add(undefined, 4)?
The expression y=4 sets the (global, in this case) variable y to 4 and returns its value, so what you are doing is equivalent to this:
y = 4
console.log(add(4))
So, the first argument is now 4 and the second is the default 2, resulting in a sum of 6, and as side-effect you created a useless global variable (or clobbered a local one).
The correct way is, as you noticed, add(undefined, 4).
If you'd like to "name" your arguments, you could use a single object as argument:
function add ({ x = 1, y = 2 } = {}) {
return x + y
}
console.log(add({ y: 4 })) // 5
This works thanks to destructuring. (The = {} allows to use add() without any argument whatsoever - without it, you'd have to write add({}).)
Note that then you need to always specify the names.
