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

Like others have said the problem is that throw is a statement and not an expression.

There is however really no need for this dichotomy. There are languages where everything is an expression (no statements) and they're not "inferior" because of this; it simplifies both syntax and semantic (e.g. you don't need separate if statements and the ternary operator ?:).

Actually this is just one of the many reasons for which Javascript (the language) kind of sucks, despite Javascript (the runtime environment) being amazing.

A simple work-around (that can be used also in other languages with a similar limitation like Python) is:

function error(x) { throw Error(x); }

then you can simply write

let x = y.parent || error("No parent");

There is some complexity in having throw as an expression for statically typed languages: what should be the static type of x() ? y() : throw(z)?; for example C++ has a very special rule for handling a throw expression in the ternary operator (the type is taken from the other branch, even if formally throw x is considered an expression of type void).