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David Rodrigues
David Rodrigues

Reputation: 12532

Why `int(0) and boolean` returns `int(0)` instead `boolean(false)`?

Specifically (I don't know if it can happen in others case), when I do int(0) && boolean with myString.length && myString === "true" when myString = "", it returns 0 instead of returns a boolean.

alert((function() {
    var myString = "";
    return myString.length && myString === "true";
})());

I know that I can fix it by do myString.length !== 0 or casting with Boolean. But I like to understand why && don't force cast the left side to boolean.

Live example: http://jsfiddle.net/uqma4ahm/1/

Upvotes: 2

Views: 37

Answers (2)

user2864740
user2864740

Reputation: 61875

The && operator yields the left hand value if it is false-y1 and the right-hand value otherwise. (Unlike some languages, it is not guaranteed to evaluate to true or false!)

TTL for A && B:

A         B     (A && B)
false-y   -     A
truth-y   -     B

To make sure that only true or false is returned, it is easy to apply (double) negation:

return !!(myString.length && myString === "true")

or, equivalently

return !(!myString.length || myString !== "true")

Here is the negation TTL which leads to deriving !!(false-y) => false and !!(truth-y) => true.

A         !A      !(!A)
false-y   true    false
truth-y   false   true

1 In JavaScript the false-y values are false 0, "", null, undefined, NaN. While everything else is truth-y.

Upvotes: 3

Pointy
Pointy

Reputation: 413712

The && operator in JavaScript returns the actual value of whichever one of its subexpression operands is last evaluated. In your case, when .length is 0, that expression is the last to be evaluated because 0 is "falsy".

The operator does perform a coercion to boolean of the subexpression value(s), but it only uses that result to determine how to proceed. The overall value of the && is thus not necessarily boolean; it will be boolean only when the last subexpression evaluated had a boolean result.

This JavaScript behavior is distinctly different from the behavior of the similar operator in C, C++, Java, etc.

Upvotes: 4

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