Does JavaScript have non-shortcircuiting boolean operators?

Question

In JavaScript

(f1() || f2()) 

won't execute f2 if f1 returns true which is usually a good thing except for when it isn't. Is there a version of || that doesn't short circuit?

Something like

var or = function(f, g){var a = f(); var b = g(); return a||b;} 

Answer 1

Nope, JavaScript is not like Java and the only logical operators are the short-circuited

https://developer.mozilla.org/en/JavaScript/Reference/Operators/Logical_Operators

Maybe this could help you:

http://cdmckay.org/blog/2010/09/09/eager-boolean-operators-in-javascript/

| a     | b     | a && b | a * b     | a || b | a + b     | |-------|-------|--------|-----------|--------|-----------| | false | false | false  | 0         | false  | 0         | | false | true  | false  | 0         | true   | 1         | | true  | false | false  | 0         | true   | 1         | | true  | true  | true   | 1         | true   | 2         |  | a     | b     | a && b | !!(a * b) | a || b | !!(a + b) | |-------|-------|--------|-----------|--------|-----------| | false | false | false  | false     | false  | false     | | false | true  | false  | false     | true   | true      | | true  | false | false  | false     | true   | true      | | true  | true  | true   | true      | true   | true      | 

Basically (a && b) is short-circuiting while !!(a + b) is not and they produce the same value.

Answer 2

You could use bit-wise OR as long as your functions return boolean values (or would that really matter?):

if (f1() | f2()) {     //... } 

I played with this here: http://jsfiddle.net/sadkinson/E9eWD/1/