Function parameter as a number for a local variable name

Question

I've just read the answers for this question about 'setting global variables inside a function' but I still have a doubt.

This maybe is very basic but can someone tell me why an error does not occur when I do this? I understand that what is passed to the function is a copy of the value of the variable i, but... why

var i = 5;

function exp(i) {
    i = 7;
    console.log(i);
}

exp(i);
//logs 7

or

function exp(i) {
  return i = 7;
  console.log(i);
}
exp(5)
//logs 7

and:

function exp() {
  return 5 = 7; //or console.log(5 = 7);
}
exp()//Uncaught ReferenceError: Invalid left-hand side in assignment

In the first example am I not making 5 = 7? Why the function logs '7'?

This all came up after I've seen this example in the wonderful JavaScript Garden about local variables:

// global scope
var foo = 1;
var bar = 2;
var i = 2;

function test(i) {
    // local scope of the function test
    i = 5;

    var foo = 3;
    bar = 4;
}
test(10);

Why test(10) which sets 10 = 5 inside the function does not make an error?

Your are not making 5=7, your setting value 7 to variabile i. If you do "return i==7" then it returns 5==7, hence false. — Franz Tesca
– Franz Tesca, Commented Aug 6, 2016 at 20:46
But if that is the case I would be changing the global variable i to 7 and that is not the case. the variable i will still be 5 (in the first example) — viery365
– viery365, Commented Aug 6, 2016 at 20:47
Possible duplicate of Is JavaScript a pass-by-reference or pass-by-value language? — castletheperson
– castletheperson, Commented Aug 6, 2016 at 20:52
In the first example you set i=7 and in fact console.log(i) will output 7... — Franz Tesca
– Franz Tesca, Commented Aug 6, 2016 at 21:00
Are you familiar with variables and assignment statements? i = i + 1 is a valid statement that increments i by 1. i = 7 changes i to 7 and forgets the previous value. — castletheperson
– castletheperson, Commented Aug 6, 2016 at 21:10

user5116395 · Accepted Answer · 2016-08-06 20:47:33Z

2

You can't do 5 = 7. 5 will always be 5.

i = 7 equals to set var i as 7

answered Aug 6, 2016 at 20:47

user5116395

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Comments

Jorawar Singh · Accepted Answer · 2016-08-06 21:19:38Z

1

your first method

var i = 5;

function exp(i) {
    i = 7;
    console.log(i);//of course it prints 7, i belong to exp scope
}
console.log(i)//it will print five
exp(i);

you are not giving new value to i variable but just to parameter you got

function exp() {
  return 5 = 7; //or console.log(5 = 7);
}

your trying to returnn 5 = 7; what's 5? In Javascript if you are not using "strict mode" then you can declare variable without var keyword, but variable name cannot start with number.

function exp() {
      return 5 + 7; //or console.log(5 + 7);//this will work
 }
 function exp(i) {
      i = 5
     return i = 7; //or console.log(5 + 7);//this will work
 }

edited Aug 6, 2016 at 21:19

answered Aug 6, 2016 at 21:08

Jorawar Singh

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1 Comment

viery365 Over a year ago

I think I understand now:) Thank you:) So, the argument of a function call gives always a value to the parameter of that function, and that parameter has to be a value, never a variable name. That's why in functions constructors function like function X(name){this.name = name}the parameter 'name' refers to the value nameand never to this.name. Now I think I understand finally:)

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