curious double var case in javascript

Question

Curious case in JavaScript where you get not exactly the most people would expect

function test() {
  var v = 2;
  var v = v++;
  console.log(v);
}
test();

Why the ++ seems ignored here? at what point the ++ operation is executed?

"v++ increments v to 3 and returns 2" what is the first, increment or return?

if it increments first, the returned value should be "3"
if it returns first 2, then after that increments, so the incremented value is 2 and v should be 3...

It's called a [postfix increment](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Increment#postfix_increment). It will assign and then increment. With `a = 0; b = a++;`, `a` will be 1 and `b` will be 0. — evolutionxbox, May 05 '22 at 14:54
@evolutionxbox thanks, I am aware, but here is rather interesting at what point is created the new variable — serge, May 05 '22 at 14:55
Yes, the second **declaration** is ignored, as the variable `v` was already declared in that scope. The *initialization* is not ignored, but the declaration is. — Pointy, May 05 '22 at 14:56
`var v = v++;` can be replaced with `v = v++;` without changing the behavior and result, if this helps you understanding the code. There is no _"the creation of the new "v""_. — jabaa, May 05 '22 at 14:57
at what point the `++` does work then? because it has any effect — serge, May 05 '22 at 14:59
`v` is initialized with `2`. `v++` increments `v` to `3` and returns `2`. `var v = v++;` assigns `2` to `v`. — jabaa, May 05 '22 at 15:00
The language specification says that the left-hand side is evaluated first, and then the right-hand side value is assigned to the left-hand side. The value of `v++` is the original value of `v`, so `v` ends up being 2. — Pointy, May 05 '22 at 15:00
I think in C and C++ such an expression might be explicitly undefined, so the runtime/compilation could do whatever it wants. In JavaScript, however, the behavior is explicitly defined. — Pointy, May 05 '22 at 15:01

jabaa · Accepted Answer · 2022-05-05T15:05:42.150

1

v is initialized with 2.
The expression v++ increments v to 3 and then returns 2.
The statement var v = v++; assigns 2 to v (overwrites the 3 in v with the 2 returned from v++).
console.log(v) prints 2

edited May 05 '22 at 15:05

answered May 05 '22 at 15:01

jabaa

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"v++ increments v to 3 and returns 2" what is the first, increment or return? – serge May 05 '22 at 15:02
1

And the assignment of 2 to `v` happens *after* the right-hand side is completely evaluated, meaning that the increment of `v` to 3 happens *before* the assignment. – Pointy May 05 '22 at 15:02
@Pointy `v++` increments `v` to `3` (first) and returns `2` (next). – jabaa May 05 '22 at 15:03
@jabaa yes, exactly – Pointy May 05 '22 at 15:03
@serge First increment, then `return`. `return` is always the last statement in a function or operation. – jabaa May 05 '22 at 15:03
@Pointy I wasn't sure if you criticize my answer or gave clarification for serge. – jabaa May 05 '22 at 15:04
I was trying to do both :) You answer is correct. – Pointy May 05 '22 at 15:05
the answer of that one says "The assignment goes first, then the incrementing" https://stackoverflow.com/a/55515365/961631 – serge May 05 '22 at 15:08
1

@serge That's wrong. First, the right-hand side is completely evaluated and then the assignment happens. – jabaa May 05 '22 at 15:09
this one really helped to understand https://stackoverflow.com/a/55515567/961631 – serge May 05 '22 at 15:10

curious double var case in javascript

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