Question

In the below code :

console.log( (false || “test”) ? “first” : “second”) );

The o/p of first part is test ( false || “test”) , so how is my final o/p first ? What happens in general when the conditional operator can’t evaluate an expression as True or False ?

Answer 1

Every value in javascript, regardless of its type, can be coerced to a boolean value.

Values that coerce to false are called “falsey”, and values that coerce to true are called “truthy”.

Here’s a fiddle demonstrating this coercion.

In this case:

(false || "test") ? "first" : "second")

(false || "test") is logically equivalent to ("test") since false || X is equivalent to X (this is called a disjunctive syllogism, if you’re interested in logics).

Any non-empty string in javascript (including the string 'false', have fun with that bug) coerces to true, so the tertiary condition evaluates to true and logs 'first'.