When is the existence of nonlocal variables checked?

Question

I am learning Python and right now I am on the topic of scopes and nonlocal statement. At some point I thought I figured it all out, but then nonlocal came and broke everyth

Answer 1

You can see what the scope of b knows about free variables (available for binding) from the scope of a, like so:

import inspect

print( "let's begin" )

def a():
    if False:
        x = 10

    def b():
        print(inspect.currentframe().f_code.co_freevars)
        nonlocal x
        x = 20

    b()

a()

Which gives:

let's begin
('x',)

If you comment out the nonlocal line, and remove the if statement with x inside, the you'll see the free variables available to b is just ().

So let's look at what bytecode instruction this generates, by putting the definition of a into IPython and then using dis.dis:

In [3]: import dis

In [4]: dis.dis(a)
  5           0 LOAD_CLOSURE             0 (x)
              2 BUILD_TUPLE              1
              4 LOAD_CONST               1 (", line 5>)
              6 LOAD_CONST               2 ('a..b')
              8 MAKE_FUNCTION            8
             10 STORE_FAST               0 (b)

 10          12 LOAD_FAST                0 (b)
             14 CALL_FUNCTION            0
             16 POP_TOP
             18 LOAD_CONST               0 (None)
             20 RETURN_VALUE

So then let's look at how LOAD_CLOSURE is processed in ceval.c.

TARGET(LOAD_CLOSURE) {
    PyObject *cell = freevars[oparg];
    Py_INCREF(cell);
    PUSH(cell);
    DISPATCH();
}

So we see it must look up x from freevars of the enclosing scope(s).

This is mentioned in the Execution Model documentation, where it says:

The nonlocal statement causes corresponding names to refer to previously bound variables in the nearest enclosing function scope. SyntaxError is raised at compile time if the given name does not exist in any enclosing function scope.