Why is this javascript expression evaluated to false?

Question

I have written a simple function that nullifies itself after its first call:

var nullify = function(){
    nullify = null;
    return 1;     
};

If I call it two times, like this:

console.log(nullify());
console.log(nullify);

Then the 1st expression will return 1, and the other will evaluate to null. All clear up to this point.

If I however do it in a single expression and wrap it in a function:

var fun = function(f){
    return f() && !f;
}

Then, for some reason:

console.log(fun(nullify));

evaluates to false, while I would expect it to be true, since 1st operand will return 1 and the other, as a negation of null, true.

The evaluation of the right-hand operand occurs when the nullify functiion has already called nullify = null, is that right? What am I missing?

The variable you are using in the nullify function is a local variable not a global variable. So you are actually creating multiple instances of a variable named nullify. — James Freund, Sep 06 '16 at 16:27

score 4 · Answer 1 · answered Sep 06 '16 at 16:23

4

What am I missing?

That your fun function is testing the variable f, which still is bound to the function you passed in, not the variable nullify (which has the value null indeed).

answered Sep 06 '16 at 16:23

Bergi

630,263
148
957
1,375

So as a result, the nullify function gets nulled, but not f? – Michał Szydłowski Sep 06 '16 at 16:26
Yes. `f` is not a reference to the `nullify` variable, it is a separate variable holding a reference to the function. – Bergi Sep 06 '16 at 16:27
@MichałSzydłowski Yes. `f` is just a copy. JS passes by value, not reference. – Oriol Sep 06 '16 at 16:27

Why is this javascript expression evaluated to false?

1 Answers1