Does Javascript Have Non-shortcircuiting Boolean Operators?

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? Somet

Solution 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.

Solution 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/

Solution 3:

JavaScript DOES have single pipe (|, bitwise OR) and single ampersand operators (&, bitwise AND) that are non-short circuiting, but again they are bitwise, not logical.

http://www.eecs.umich.edu/~bartlett/jsops.html

Solution 4:

If you need f2() to run regardless of whether or not f1() is true or false, you should simply be calling it, returning a boolean variable, and using that in your conditional. That is, use: if (f1() || f2IsTrue)

Otherwise, use single bar or single ampersand as suggested by GregC.