How can javascript have lexical scope if scope is assigned at runtime?

Question

Im confused about javascript having lexical scope.

Lexical scope: Lexical scoping means whatever variables are in scope where you define a function from (as opposed to when you call it) are in scope in the function.

However in JS its: scope isn't assigned when the function is defined and compiled rather it's assigned at runtime i.e. when the function is called

In the below example: If JS had lexical scope, I would get an error. However I do not because of hoisting and at run time js engine checks for x.

function f() {
 console.log(x);
}
const x = 10;
f(); // 10

Can someone explain with an example how JS can have lexical scope? Thanks!

score 2 · Answer 1 · answered Nov 04 '20 at 18:26

However in JS its: scope isn't assigned when the function is defined and compiled rather it's assigned at runtime i.e. when the function is called

I think you are confusing yourself with this description of scope in JS. A more accurate description of scope in JS is given by this article (emphasis added):

Because functions are also defined at this time (lexing phase), we can say that lexical scope is based on where variables and blocks of scope exist at author time, and thus are locked down at the end of the lexing phase. Scope is not defined at runtime, rather it can be accessed at runtime.

So, in your example:

function f() { 
  console.log(x); 
} 
const x = 10; 
f(); // 10

What is happening is that the scope chain of function f consists of:

its local scope
the outer, global scope

and this "scope chain" is defined at author time (i.e. lexical scope).

At author time, the constant x is defined within the global scope, and when f is called, x is available within the global scope. That x is not defined when the function f is declared (but NOT called) is irrelevant for this discussion.

Also, you are incorrect when you mention "hoisting" regarding your code example: variables declared with const are NOT hoisted.

If we alter your example, we will get a Reference Error, since the constant x has not been initialized in the global scope (i.e. on author time), when the function f is called (and function f looks through its scope chain, which has been defined on author time):

function f() { 
  console.log(x); 
} 
f(); // 10
const x = 10;

score 0 · Answer 2 · answered Jan 06 '20 at 11:35

See below example, everything outside function f is compiled and vars are hoisted. But the contents inside function f are not hoisted/parsed until it's called. And the local variables inside f remain accessible within the function (i.e. the lexical scope of function).

function f() {
  console.log(x);
  var a = 20;
}

const x = 10;

f(); // 10


console.log(a);

How can javascript have lexical scope if scope is assigned at runtime?

2 Answers2