python - 告诉函数如何接收参数？

Question

我想知道在 cpython 中是否可以进行以下自省(introspection):

>>> def potato(x=69):
...     if x == 69 and ???:
...         print '69 was taken from argument defaults'
...     if x == 69 and ???:
...         print '69 was passed as positional arg'
...     if x == 69 and ???:
...         print '69 was passed as kwarg'
...
>>> potato()
69 was taken from argument defaults
>>> potato(69)
69 was passed as positional arg
>>> potato(x=69)
69 was passed as kwarg

我对 python2 和 python3 的答案都感兴趣，如果它们不同的话。

任何涉及inspect、traceback、pdb、sys._getframe 等的黑魔法在这里都是允许的。当然，修改函数的 argspec 是不允许的。

Answer 1

虽然框架有一个名为 code_context 的字符串，但它看起来不像 inspect 可以直接提供此信息，它为您提供了调用该函数的源代码行。问题是人们必须重写一个小的解析器才能理解它。

这是一个更简单的解决方案，它基于您要检查的函数的包装器。它不会改变 arg 规范，arg 验证也不会改变:

import inspect

def track_args(func):
    def tracker(*args, **kwargs):
        r = func(*args, **kwargs)
        for arg_num,arg_name in enumerate(inspect.getargspec(func)[0]):
            if arg_name in kwargs:
                print "%s was provided as keyword arg" % arg_name
            elif arg_num < len(args):
                print "%s was provided as positional arg" % arg_name
            else:
                print "%s was provided by default value" % arg_name
        return r
    return tracker

@track_args
def f(a, b, c=30):
    print "I'm f()"

f(20, b=10)
f(20)

具有有效参数的结果:

I'm f()
a was provided as positional arg
b was provided as keyword arg
c was provided by default value

参数无效的结果:

Traceback (most recent call last):
  File "test.py", line 21, in <module>
    f(20)
  File "test.py", line 5, in tracker
    r = func(*args, **kwargs)
TypeError: f() takes at least 2 arguments (1 given)