Coffee Existential Operator different compliations [duplicate] - coffeescript

Question

This question already has an answer here:
How does CoffeeScript's existential operator work?
(1 answer)
Closed 8 years ago.
Similar to this topic:
CoffeeScript Existential Operator and this
The coffeescript, I'm having a problem using the elvis operator:
foo = ->
y = foo()
console.log 'y is null' unless y?
console.log 'x is null' unless x?
Compiles to:
var foo, y;
foo = function() {};
y = foo();
if (y == null) {
console.log('y is null');
}
if (typeof x === "undefined" || x === null) {
console.log('x is null');
}
Output:
y is null
x is null
So the problem is that since y is assigned earlier that coffee takes the shortcut and assumes that y cannot be undefined. However, it is valid to return undefined from a function.
Is there a "safer" way to check that y is also not undefined?
UPDATED
Clarified Examples and Explanation:
From the comments, in the first if statement (y == null) is using double equal instead of ( x === null) triple equal, as in the second if statement. Clever.

Answer 1

? operator always checks if the value is neither null nor undefined.
y == null is absolutely correct way to check that the value of y is either null or undefined in JavaScript.
For example, the following CofeeScript code checks only for null values:
do_something() if y is null
which compiles to
if (y === null) do_something();
So, while y? (y == null in JS) checks for both null and undefined, y isnt null (y !== null in JS) checks only for null.
Check this answer for more detailed information.
See this answer for more info about equality checks in JavaScript.