AS3 && || operators on instructions

Question

what does the operators evaluate on instructions?

like:

var flag:Boolean=true;
flag && trace("1") && trace("2") && trace("3");

output: 1 2 3

var flag:Boolean=true;
flag && trace("1") || trace("2") || trace("3");

output 1

The or operator breaks the sentence under idk what circumstances...

Edit: ok, now i see how the sentence works with the operators, but is any instruction that doesn't have a return value valuated as true?

And it appears u can't evaluate any instruction u want, like:

private function any():void{
true && return;
}

that will throw a compilation error.

Edit: In this case, it behaves different to the second example:

true && one() || two() || three();

the functions

private function one():void{
    trace("1");
}
private function two():void{
    trace("2");
}
private function three():void{
    trace("3");
}

output: 1 2 3

Edit: Assigning values:

var a:int;
(a=1) || trace("1");
(a=2) && trace("2");

output: 2

Edit: || Returns expression1 if it is true or can be converted to true, and expression2 otherwise

&& "Returns expression1 if it is false or can be converted to false, and expression2 otherwise. Examples of values that can be converted to false are 0, NaN, null, and undefined".

http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/operators.html#logical_OR

http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/operators.html#logical_AND

Ahe · Accepted Answer · 2012-11-02 21:27:54Z

6

This is called short-circuit evaluation. Basically the and/or expression evaluation is stopped when its result can be determimed for sure.

For example when evaluating and (&&) evaluation can be stopped at the moment first false is found. false && true && true && true... is always false, no matter how many true-values you add to end. Same goes for or and true-value.

edited Nov 2, 2012 at 21:27

answered Nov 2, 2012 at 21:22

Ahe

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1 Comment

Ziul Over a year ago

oh, that really makes sense, and explains why the sentence breaks, but any instruction will always return true??

zzzzBov · Accepted Answer · 2012-11-03 22:16:46Z

5

ActionScript 3 is an ECMAScript variant, which means it's based off the same standard as JavaScript.

Because of this && and || are not similar to other C-style languages.

a || b returns a if a is truthy, otherwise it returns b.

a && b returns a if a is falsey, otherwise it returns b.

This makes || essentially equivalent to a ? a : b, and && essentially equivalent to !a ? a : b.

They also have a short-circuit evaluation mechanism. If the first argument is to be returned, the second argument is not executed.

edited Nov 3, 2012 at 22:16

answered Nov 2, 2012 at 21:24

zzzzBov

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4 Comments

Ziul Over a year ago

could u provide a little example on this: "|| essentially equivalent to a ? a : b"

zzzzBov Over a year ago

@wvxvw, if you're using strongly typed objects, you'll receive strongly typed values. If you use dynamically typed objects (i.e. var a:*, or simply var a) you'll receive the dynamically typed value.

zzzzBov Over a year ago

@wvxvw, simply put, you're mistaken. || is often used for its coalescing behavior when setting objects. a = a || {} works because a will be truthy if set, and falsey if undefined. I will update my answer with links to the documentation, please take some time to review it.

zzzzBov Over a year ago

@wvxvw, no problem. I was beginning to worry that I'd gotten it wrong for AS3, as it's been quite a while since I've actually written any flash code.

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