Why doesn't logical OR work with error throwing in JavaScript?

Question

This is a pretty common and useful practice:

// default via value
var un = undefined
var v1 = un || 1

// default via a function call
var myval = () => 1
var          


        

Answer 1

You could move the throwing of the exception into a function, because throw is a statement of control flow, and not an expression:

An expression is any valid unit of code that resolves to a value.

const throwError = function (e) { throw new Error(e); };

var un = undefined,
    v3 = un || throwError('un is not set!');