为什么JavaScript的“ in”运算符会错误地返回false？

Question

You're able to replicate by so:

var test = {'var1': 'bacon'};

"var1" in test;             // Returns true  - Correct
!"var1" in test;            // Returns false - Correct
"nonexistant" in test;      // Returns false - Correct
!"nonexistant" in test;     // Returns false - Incorrect - This should be true.. should it not?

Answer 1

The in operator binds fairly loosely. It's generally a good idea to parenthesize in subexpressions.

Thus, !"var1" in test is parsed as (!"var1") in test for example.