Question

>>> function () {}
SyntaxError

>>> f = function() {}
f = function() {}

>>> new function() {}
>>> Object

>>> new Function()
function anonymous() {}

>>> Function()
function anonymous() {}


>>> f = (function() { a = 10; return function() {console.log(a);} })();
>>> f()
10
undefined

>>> f = (function() { a = 10; return new Function('console.log(a)'); })();
>>> f()
undefined

Thus, I have two questions:

1. Why is the Function constructor returning a new function even without the new operator?

2. Are functions created with the Function constructors NOT closures?

Answer 1

Why is the Function constructor returning a new function even without the new operator?

Calling Function() is the same as calling new Function():

When Function is called as a function rather than as a constructor, it creates and initialises a new Function object. Thus the function call Function(…) is equivalent to the object creation expression new Function(…) with the same arguments.

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Are functions created with the Function constructors NOT closures?

Yes, the scope of the function is set to the global environment, not to the scope of the lexical environment. See http://es5.github.com/#x15.3.2.1, step 11:

11. Return a new Function object created as specified in 13.2 passing P as the FormalParameterList and body as the FunctionBody. Pass in the Global Environment as the Scope parameter and strict as the Strict flag.

That means using the Function constructor is as if you declared a function in global scope, regarding the scope it can access (not regarding where the function is visible).

This is different from using function declarations/expressions, where the scope is based on the current exectution context (http://es5.github.com/#x13):

1. Let funcEnv be the result of calling NewDeclarativeEnvironment passing the running execution context’s Lexical Environment as the argument