javascript - Can (a== 1 && a ==2 && a==3) ever evaluate to true?

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Is
it ever possible that (a== 1 && a ==2 && a==3)
could evaluate to true in
JavaScript?

This is an interview question asked
by a major tech company. It happened two weeks back, but I'm still trying to find the
answer. I know we never write such code in our day-to-day job, but I'm
curious.

Answer

If you take advantage of href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Equality_comparisons_and_sameness#Loose_equality_using"
rel="noreferrer">how == works, you could simply
create an object with a custom toString (or
valueOf) function that changes what it returns each time it is
used such that it satisfies all three conditions.

class="snippet" data-lang="js" data-hide="false" data-console="true"
data-babel="false">

class="snippet-code-js lang-js prettyprint-override">const a =
{
i: 1,

toString: function () {
return
a.i++;
}
}

if(a == 1 && a == 2
&& a == 3) {
console.log('Hello
World!');
}

/>

The reason this works is due to the use of the loose
equality operator. When using loose equality, if one of the operands is of a different
type than the other, the engine will attempt to convert one to the other. In the case of
an object on the left and a number on the right, it will attempt to convert the object
to a number by first calling valueOf if it is callable, and
failing that, it will call toString. I used
toString in this case simply because it's what came to mind,
valueOf would make more sense. If I instead returned a string
from toString, the engine would have then attempted to convert
the string to a number giving us the same end result, though with a slightly longer
path.