I would like to use the CoffeeScript existential operator to check some object properties for undefined. However, I encountered a little problem.
Code like this:
console.log test if test?
Compiles to:
if (typeof test !== "undefined" && test !== null) console.log(test);
Which is the behavior I would like to see. However, when I try using it against object properties, like this:
console.log test.test if test.test?
I get something like that:
if (test.test != null) console.log(test.test);
Which desn't look like a check against undefined at all. The only way I could have achieved the same (1:1) behavior as using it for objects was by using a larger check:
console.log test.test if typeof test.test != "undefined" and test.test != null
The question is - am I doing something wrong? Or is the compiled code what is enough to check for existence of a property (a null check with type conversion)?
This is a common point of confusion with the existential operator: Sometimes
x?
compiles to
typeof test !== "undefined" && test !== null
and other times it just compiles to
x != null
The two are equivalent, because x != null will be false when x is either null or undefined. So x != null is a more compact way of expressing (x !== undefined && x !== null). The reason the typeof compilation occurs is that the compiler thinks x may not have been defined at all, in which case doing an equality test would trigger ReferenceError: x is not defined.
In your particular case, test.test may have the value undefined, but you can't get a ReferenceError by referring to an undefined property on an existing object, so the compiler opts for the shorter output.