Why does Chrome debugger think closed local variable is undefined?

Question

With this code:

function baz() {
  var x = \"foo\";

  function bar() {
    debugger;
  };
  bar();
}
baz();

I get this unexpected result:<

Answer 1

Like @Louis said it caused by v8 optimizations. You can traverse Call stack to frame where this variable is visible:

Or replace debugger with

eval('debugger');

eval will deopt current chunk