why in JavaScript's: obj = new Boolean(false) , (obj && true) is true, and (obj || false) is false? -


this question has answer here:

to see result, open console in google chrome, following:

obj = new boolean(false) "obj && true: " + (obj && true) "obj || true: " + (obj || false)

and: 

(obj && true) == true   // true (obj || false) == true   // false

and why

(obj || false)

returns boolean object, instead of boolean value?

(hmm.. put summary answer below)

there 2 concepts need considered here:

obj = new boolean(false)

creates object, values false. object considered truthy, it's value (which tostring() or valueof()) of course boolean value false.

(x||y)

returns first truthy value (or if none present, last falsy value) and

(x&&y)

returns first falsy value (or if none present, last truthy value).

so (obj||false) returns boolean object, (obj&&true) returns second (true) value.

the further preceeding depends on context of expression.

"obj && true: " + (obj && true)

demands string context, tostring() called on boolean object, returning it's value false (while object truthy!).

furthermore,

(obj && true) == true compares true == true of course true. however,

(obj || true) == true lot of type coercion §11.9.3 , compares

toprimitive(obj) == tonumber(true) (§9.1 , §9.3) results in nan == 1 yields false.

the results more predictable if use strict equality operator §11.9.6.


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