CODEWITHSUNDEEP
javascript
Donato
Donato

Reputation: 2777

post-increment operator with shorthand assignment

I have a simply array and counter:

var data = [1];
var i = 0;

The shortcut assignment produces 2:

data[i++] *= 2
// 2

I was expecting 3. data[i++] is multipled with 2, so 1 * 2 is 2, and then that is assigned to data[i++], which now becomes 2, and then after statement evaluated the side effect of ++ causes i to be 3.

The following also gives me unexpected result. It produces NaN oddly enough.

var data [1];
var i = 0;
data[i++] = data[i++] * 2;
// NaN

I was expecting 3 again. data[i++] first evaluates to 1, and then is multipled with 2, and then that 2 value is assigned to i in data[i++], which is then incremented after statement completes, causing it to be 3.

What am I missing here?

Upvotes: 1

Views: 192

Answers (2)

Amadan
Amadan

Reputation: 198314

  • var data [1]; is not a valid JavaScript. Did you mean var data = [1];?

  • data[i++] *= 2 is evaluated as follows:

    • i++, as the innermost expression resolves first: its value is i (i.e. 0), and i increments afterwards to 1.

    • data[0] is looked up, and multiplied by two; since data[0] is 1, data[0] gets assigned the value of 1 * 2, i.e. 2.

    • The value of the outermost expression is returned: 2. ++ increments only what it was applied to (i), and not the whole expression.

  • data[i++] = data[i++] * 2 evaluates as follows:

    • The first i++ evaluates to 0 and modifies i to 1, as before.

    • The second i++ evaluates to 1 and modifies i to 2

    • The expression then evaluates as data[0] = data[1] * 2. data[1] is undefined, and undefined * 2 is not a number (NaN).

  • In general, it is strongly recommended to avoid having two increment/decrement operators in the same expression. Different languages (and indeed, different compilers of the same language) have wildly different ideas of what should be done. In many languages, it is declared "undefined behaviour" in language specification.

Upvotes: 5

Amar Syla
Amar Syla

Reputation: 3653

Instead of i++, use ++i. In your case, you're first returning i, then incrementing it, while you're looking for an increment, and return it after.

Upvotes: 1

Related Questions