Why is the value in this js code not changed?

Question

I was looking at this javascript code:

var a = {i: 0};
var b = 0;

function change(a, b) {
  a.i++;
  b++;
}
change(a, b);

console.log(a, b);

You would expect to be logged: {i:1} 1 . It actually returns: {i:1} 0? Why is b=0 and not b=1?

score 4 · Accepted Answer · answered Jul 01 '18 at 07:41

4

Because b is a parameter inside change - inside change, b refers to the local variable, and when the local variable is reassigned, that has no effect on the b outside.

Either always refer to b as a global variable:

var a = {i: 0};
var b = 0;

function change(a) {
  a.i++;
  b++;
}
change(a);

console.log(a, b);

Or return b from change, and assign the outer b to the result of change:

var a = {i: 0};
var b = 0;

function change(a, b) {
  a.i++;
  b++;
  return b;
}
b = change(a, b);

console.log(a, b);

(if you need to return and reassign multiple variables, return an object or array:

var a = {i: 0};
var b = 0;
var c = 0;

function change(a, b, c) {
  a.i++;
  b++;
  c++;
  return { b, c };
}
({b, c} = change(a, b, c));

console.log(a, b, c);

)

answered Jul 01 '18 at 07:41

CertainPerformance

356,069
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309
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The a variable has changed though? – bier hier Jul 01 '18 at 07:45
... or pass b inside an object – Dominic Jul 01 '18 at 07:45
@bierhier `a` is an *object* - when `a` is passed, you're only passing a *reference* to the same `a` object in memory, so when `a` is mutated inside the function, it's changed on the outside as well. – CertainPerformance Jul 01 '18 at 07:47

score 1 · Answer 2 · answered Jul 01 '18 at 07:44

1

You're shadowing your variable a in your function header.

function change() {
    a.i++;
    b++;
  }

will work.

answered Jul 01 '18 at 07:44

Moritz Roessler

8,542
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51

Why is the value in this js code not changed?

2 Answers2