In JavaScript, functions can have properties. Like this:
var f = function(){ console.log("Hello!"); }
f.x = "Whoohoo";
How do I retrieve x from code within f() that might be called long after the variable f goes out of scope?
4 Answers
You could use the property accessor, like the assignment.
var f = function() { console.log("Hello! " + f.x); }
f.x = "Whoohoo";
console.log(f.x);
f();For stable access, you could use a named function
var f = function foo() { console.log("Hello! " + foo.x); }
// ^^^ >>>>>>>>>>>>>>>>>>>>>>>>>>> ^^^
f.x = "Whoohoo";
console.log(f.x);
f();5 Comments
Seva Alekseyev
You misunderstood; I want to get the property from code within the function itself, not from the code that surrounds it.
Nina Scholz
@SevaAlekseyev, it is stil the property accessor. for a static reference, i suggest to use a named function.
Seva Alekseyev
Works. It wasn't immediately obvious to me that the closure logic would capture the function variable, since it hasn't been assigned at the time of function creation.
Damon
Nice - using a named function would be preferable to the function referencing
f
Bergi
@SevaAlekseyev Closures do capture bindings (variables), not the values that they contain when the closure is created.
If you need some more robust (but also more verbose):
var f = (function(){
var func = () => {
console.log(func.a)
}
return func
})()
f.a = 'Whoohoo'
// trying to break it
var a = f
f = 'somethingelse'
a()
Comments
If what you're looking for is to simply access that property from within your function f, you simply reference the function object (either by the variable name if it's a function expression, or by the function name if it's a function declaration).
For example:
var f = function(){ return f.x; };
f.x = "Whoohoo";
// f() => "Whoohoo"
var b = function(){ return f.x; ];
// b() => "Whoohoo"
However, your mention of calling f "long after the variable f goes out of scope" contradicts the lexical scoping of JavaScript. You cannot call the function f unless you do so from the same scope where the function was defined, or from a child scope in reference to where the function was defined. If you are "out of scope", invoking f will throw a reference error.
1 Comment
Seva Alekseyev
setTimeout(f, 1);Let's get a little convoluted. f is already the name of the function by the expression that defines it. And a function is an object by definition in JS. So you can access it and it's properties (such as x) through the f keyword like f.x. Yet, since f is a function, you may use the it as a constructor too and do silly things like...
var f = function(){ console.log("Hello!"); };
f.x = "Whoohoo";
f.prototype.thingy = function(){
console.log(this.constructor.x);
};
var a = new f();
a.thingy();... and from this point on lets get a little more convoluted. You might event do;
function F(){
F.prototype.thingy = function(){
return this.constructor.x;
};
var obj = Object.create(F.prototype);
console.log("Hello!");
return obj.thingy();
}
F.x = "Whoohoo";
console.log(F());So you can get the value of a property of the function object F "from within itself" by taking an object instantiated by itself under a closure.
f, how is it possible for it to be "out of scope"? Just usef.xin the body.