Why plus operator in brackets works differently?

Question

Could you please explain why the results are different?

({} + {}) // = "[object Object][object Object]"
{} + {} // = NaN

I understand that in both cases objects are converted to strings, but why in the second case the result is converted to a number?

It's not the plus operator that is acting different, it's those curly braces that are once an object literal and once something else — Bergi, Jul 23 '15 at 10:07
@Ezee: Read the linked post. It's not an expression, but an empty block statement. more details [here](http://stackoverflow.com/q/26347326/1048572) — Bergi, Jul 23 '15 at 10:10
@Bergi: That dupe doesn't answer why `({} + {})` returns something else than `{} + {}` does. — Cerbrus, Jul 23 '15 at 10:14
Well, the answer is trailing semicolons indicated by @ÁlvaroG.Vicario — Bhojendra Rauniyar, Jul 23 '15 at 10:15

score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

From:

http://www.2ality.com/2012/01/object-plus-object.html

The problem is that JavaScript interprets the first {} as an empty code block and ignores it. The NaN is therefore computed by evaluating +{} (plus followed by the second {}). The plus you see here is not the binary addition operator, but a unary prefix operator that converts its operand to a number, in the same manner as Number()

...

Why is the first {} interpreted as a code block? Because the complete input is parsed as a statement and curly braces at the beginning of a statement are interpreted as starting a code block. Hence, you can fix things by forcing the input to be parsed as an expression:

({} + {})

'[object Object][object Object]'

Why plus operator in brackets works differently?

1 Answers1