为什么`int（0）和boolean`返回`int（0）`而不是`boolean（false）`？

Question

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/

Answer 1

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.

Answer 2

The && operator yields the left hand value if it is false-y ¹ 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.