How does the function run even though i have not called it

Question

function sayHello(name) {
var text = 'Hello ' + name;
var say = function() { console.log(text); }
say();
}
var x = sayHello('Joe');

Answer 1

3) when a function is called before declaration of the function
// it uses it as the main function

and vice versa if uses as the secondary /sub function

//if multiple declaration of sayNumber()

then

if there is only one function with saynumber();

then depends on only one function declared

Answer 2

when the program gets the first name in anywhere inside after declaration of a function,

the compiler knows program is calling the function

example:- declared anywhere

Answer 3

Then how do I assign sayHello('Joe') to X without calling the function

Answer 4

use different name , do not use the name of the function

Answer 5

Can you give me an example

Answer 6

var y= // anything else

Answer 7

@Rajeeb

function say667() {
var num = 42;
var say = function() { console.log(num); }
num++;
return say;
}
var sayNumber = say667();
sayNumber();

//Output: 43

function say667() {
var num = 42;
var say = function() { console.log(num); }
num++;
return say;
}
var sayNumber = say667();

//Output: Blank

1. Why does num = 43(and not 42) although ++ operator is after the function say()?
2. Why is there no output when I removed sayNumber();
3. What does sayNumber(); exactly do
4. Why is there a necessity to have return say;

I am a newbie and i have a lot of questions, please bear with me! It will really make my day if you answer these questions for me

Answer 8

1) multiple declaration of functions sayNumber()

, where one function will affect the other

the variable num gets affected twice in the code

// num= 43 not 42

Answer 9

2) // no output

no call for the function second time

at the end of second function saynumber

Answer 10

4) return necessity

to return a value to the function
no necessary unless you provide arguments to the functions

Answer 11

Can you elaborate Number 1. Thank You