Javascript: assigning the second optional argument without assigning the first

Question

I'm trying to create a function that has multiple optional arguments. I am then trying to only assign the second argument without having to also assign the first one.

function foo(bar, zar) {
  bar = bar || 5;
  zar = zar || 2;

  return bar * zar;
}

foo()      //10
foo(7)     //14
foo(7, 3)  //21
foo(zar=5) //10 <-- why is this not 25?

What is the best way to do this?

Answer 1

IMHO, the easiest way to handle this kind of situation is to instead use an object for the parameter, with default field values via the ES6 syntax.

 function foo({ bar = 5, zar = 2 } = {}) { console.log(bar * zar); } foo() //10 foo({ bar: 7 }) //14 foo({ bar: 7, zar: 3 }) //21 foo({ zar: 5 }) //25 

For reference, see: Default parameters and Object destructuring .

Answer 2

You could pass undefined for whatever parameters you would want to use as defaults.

foo(undefined, 5);

Also, if you are free to use ES6, you can simplify your function a little and declare your optional parameters like so:

function foo(bar = 5, zar = 2) { }

Answer 3

Use void 0 instead of the arguments you don't want to supply.

For example: foo(void 0,zar=5);

Any unfilled argument will be undefined .

You cannot pass the second parameter without passing the first, but you can pass undefined . In the function, you can check for undefine d and replace with a default value.

Alternately, you can use an object argument to imitate keyword arguments.