javascript variable as function [duplicate]

Question

This question already has answers here:

Javascript function scoping and hoisting (16 answers)

Closed 1 year ago.

If a javascript variable is called as function, then what is the working process?

let a=5;
function show(){
 a=10;
 function a(){};
 return a;
}
let res=show();
console.log(a,res); //output:5 10

Can anyone please explain why it is showing 5 for a, what is the meaning of function a(){}?

• Carsten Massmann: Your function show() assigns the value 10 to the global variable a and defines the function a() in its own scope. It appears that the global variable a and not the function a() is returned at the end of the function.
• Guy Incognito: @CarstenMassmann No, it's the opposite. The definition of functiona() is hoisted at the top of the function show() as a local variable and then overwritten with the number 10. The function show() then returns the local variable a. (If it changed the global variable a then console.log(a, res) would not print 5 as its value.)
• Carsten Massmann: The function definition of a() causes a local instance of a to be created in show(). This (creation of functions) always happens before any other code in that function is executed. Afterwards a gets a new value: 10, and this is eventually returned at the end of the function. Yes, @GuyIncognito, I just read your comment and I agree - "hoisting" is the magic word here! ;-)

Answer 1

In JavaScript, all function declarations are hoisted to the top of the scope. Function declarations themselves create a scope, so in this case, the code is really interpreted as:

let a=5;
function show(){
 function a(){};
 a=10;
 return a;
}
let res=show();

This means that the outer a reference never changes. A related SO question: javascript scope of function declarations