Can !a!=!!b^!!-!a||!+!a|!c return anything other than 1?

Question

I was playing the Javascript game with somebody and we were having fun making ridiculous and absurd expressions to make our inputs get a particular output.

This little charming one

!a!=!!b^!!-!a||!+!a|!c 

always seemed to return 1. I tried to reason it out, but I gave up after losing track of all the !s.

Are there any values for a, b, and c which do not return 1? If not, why does it always return 1?

Answer 1

Short answer, yes. a = false, b = false, c = true is a counter-example because your equation is identical to (!!a || !!b || !c).

Long answer:

!a!=!!b^!!-!a||!+!a|!c 

is

(((!a) != (!!b)) ^ (!!(-!a))) || ((!+!a)|!c) 

which reduces to

((Boolean(a) == Boolean(b)) ^ (!a)) || (Boolean(a) | !c) 

so all of a, b and c are only dealt with as truthy/falsey values and the result must be a 1 or 0 since | and ^ both coerce booleans to numbers.

So obviously (from inspection of the right of the ||) if either a is truthy or c is falsey, you get 1.

If a is falsey and c is truthy, you have two possibilities,

1. b is truthy in which case the ^ clause is 1 so the right of the || is never reached.
2. b is falsey, in which case the ^ clause is 0 so the right of the || dominates to produce 0.

Answer 2

How about this:

var a = undefined, b=undefined, c=!a alert(!a!=!!b^!!-!a||!+!a|!c) // Output: 0 

Live demo.