The following code demonstrates an immediately invoked function expression used as the right-hand value for a default value in a destructuring assignment expression.
const { foo = (function bar() { return 'bam'; }()) } = {};
Function expression bar is not visible outside of the expression.
Does this mean JavaScript has expression scope?
If so, does this mean that the expression has its own conceptual LexicalEnvironment?
The function expression bar is not visible because it's a function expression, even if a named one. It does not create any symbol for that function in the symbol table.
function bar() {}
This is a function declaration which creates a symbol bar.
(function bar() {});
This is a function expression which does not result in anything. It's a named function expression so you'd see a function name pop up in stack traces and the function can refer to itself by name, but it's not a function declaration.
bar, does not impact the EnvironmentRecord of the outer LexicalEnvironment ("symbol table" in your answer). But given that the function expression will be given a LexicalEnvironment of its own when it is invoked, does that mean JS has expression scope or not? I can't figure it out. I would guess no?
Yes and yes.
The official spec for JavaScript even has a section talking about lexical environments.
At a high level, basically any closure or block will define a new level of scope, where functions aren't available outside of them. This is used to help provide scoping and do things like emulate "private" variables.
AssignmentProperty : IdentifierReference Initializer