python - 如何从 Python shell 调用内部函数？

Question

我有一些代码（我不能轻易修改），格式如下：

def foo(x):
   do_other_stuff_that_I_do_not_want_to_do()
   def bar():
      "do something"
      str(x)
   bar()

我想bar()直接从 Python shell 调用 。我不介意使用co_globals, 或其他内部位。我觉得这可能是不可能的；是吗？

Answer 1

It is impossible to get at the inner function object with the code as you've stated it -- said object is only created (by the def statement) when the outer function runs (i.e., when it gets called).

As an aside, note that outer functions like foo are often coded to return the inner function as their result (e.g. by changing bar() to return bar) as the last line, exactly to work as "function factories" (often, as "closure factories") rather than keep the very existence of an internal function as a kind of private, secret implementation detail; but your coding picks the latter route instead.

Edit...:

It is possible to get at the code object for the inner function, however:

>>> def outer():
...   def inner(): return 'boo'
...   print inner()
... 
>>> eval(outer.func_code.co_consts[1])
'boo'
>>> 

However, in general, to make a code object into a callable function requires considerable work (in this special case an eval suffices because the inner function has no arguments nor any nonlocal variables, but of course that's not anywhere even close to the general case... in which you do have to supply such niceties as bindings for nonlocals!-)