javascript - execution context of function

Question

Why does z() execution context not override global x variable?

var x = 10;

function z(){
  var x = x = 20;
}
z();
console.log(x); // why is 10 printed? Shouldn’t it be 20.

var a = b = c = 0;

It means b and c are being declared as globals, not locals as intended.

For example -

var y = 10;

function z(){
  var x = y = 20; // global y is overridden
}
z();
console.log(y); // value is 20

Going by above logic, x = x = 20 in z() means x is global which overrides the local x variable but still global value of x is 10

No, as the x in the function is local to the function, your log use the global, which is 10 — Asons
– Asons, Commented Nov 10, 2017 at 18:06
In the context of var x = 10. Here X is a global variable, right ? Now, inside your z function, its a LOCAL Variable. if you remove the var from your function. It will print 20. Because it will be refering to the global one and not the local variable. — PlayHardGoPro
– PlayHardGoPro, Commented Nov 10, 2017 at 18:12
@PlayHardGoPro It's not that simple. According to the operator precedence, OP's code should actually print 20. See Ori Drori's answer. — Teemu
– Teemu, Commented Nov 10, 2017 at 18:14
@AlperFıratKaya It's not that simple. Read this if you really want to understand scope and hoisting(davidshariff.com/blog/…). — Vivek Kumar
– Vivek Kumar, Commented Nov 10, 2017 at 18:34
@VivekKumar Notice, that in the edited question, var a = b = c = 0; only a is declared, other variables refer to earlier declared variables or to an outer scope. If you want a single line declaration, you use comma to separate the declarees. — Teemu
– Teemu, Commented Nov 10, 2017 at 19:00

Ori Drori · Accepted Answer · 2017-11-10 18:06:41Z

4

The internal x declaration is hoisted to the top of the function, and overshadows the x of the external scope. Your code actually does this:

var x = 10;

function z(){
  var x;
  x = x = 20;
}
z();
console.log(x); // why is 10 printed? Shouldn’t it be 20.

answered Nov 10, 2017 at 18:06

Ori Drori

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3 Comments

This is a special case, which might be a good addition to the accepted answer in the linked dup ..?

@Teemu - it's more related related to hoisting than scoping.

Well, it messes up the expected scoping (considering the operator precedence), if you're not aware of hoisting.