python - 从 FrameInfo 或 python 中的框架获取函数签名

Question

在用 python 编写自己的异常 Hook 时，我想到了使用 inspect - 为我自己提供有关如何调用该函数的更多信息的模块。

这意味着函数的签名以及传递给它的参数。

import inspect

frame_infos = inspect.trace() # get the FrameInfos

for f_idx, f_info in enumerate(frame_infos):
    frame_dict.update(f_info.frame.f_locals) # update namespace with deeper frame levels
    #Output basic Error-Information
    print(f'  File "{f_info.filename}", line {f_info.lineno}, in {f_info.function}')
    for line in f_info.code_context:
        print(f'    {line.strip()}')

    ########################################################
    # show signature and arguments 1 level deeper
    if f_idx+1 < len(frame_infos): 
        func_name = frame_infos[f_idx+1].function #name of the function
        try:
            func_ref = frame_dict[func_name] # look up in namespace
            sig = inspect.signature(func_ref) # call signature for function_reference
        except KeyError: sig = '(signature unknown)'
        print(f'    {func_name} {sig}\n')
        print(f'    {frame_infos[f_idx+1].frame.f_locals}\n')

这对于像这样的基本示例来说效果很好:

def test1 ( x: int, y: tuple = 0 )->list: # the types obviously dont match
    return test2(y, b=x, help=0)

def test2 ( a, *args, b, **kwargs ):
    return a + b / 0

try: 
    test1(5) 
except: ...

输出:

File "C:/test/errorHandler.py", line 136, in <module>
    test1(5)
    test1 (x:int, y:tuple=0) -> list
    {'y': 0, 'x': 5}
File "C:/test/errorHandler.py", line 130, in test1
    return test2(y, b=x, help=0)
    test2 (a, *args, b, **kwargs)
    {'kwargs': {'help': 0}, 'args': (), 'b': 5, 'a': 0}
File "C:/test/errorHandler.py", line 133, in test2
    return a + b / 0

但是，一旦您离开 1 个文件，您就无法将函数名称映射到基本命名空间。

file1: import file2; try: file2.foo() except: ...
file2: import file3; def foo(): file3.foo()
file3: def foo(): return 0/0

非常重要，我正在寻找一种从 <function foo at 0x000002F4A43ACD08> 获取函数(如 FrameInfo )的方法或 frame -object，但我看到的唯一信息是名称、文件和行。
(我不喜欢通过在特定行查看源文件来获取签名的想法。)

迄今为止最好的引用是 Inspect -documentation ，但我还没有找到有用的东西。

Answer 1

基于 this answer通过 jsbueno ，我找到了恢复签名的解决方案。

使用gc(垃圾收集器)函数get_referrers()您可以搜索直接引用特定对象的所有对象。

通过框架的 f_code 提供的代码对象，您可以使用此函数来查找框架本身以及函数。

code_obj = frame.f_code
import gc #garbage collector
print(gc.get_referrers(code_obj))
# [<function foo at 0x0000020F758F4EA0>, <frame object at 0x0000020F75618CF8>]

所以，只要找到真正的函数就可以了:

# find the object that has __code__ and is actally the object with that specific code    
[obj for obj in garbage_collector.get_referrers(code_obj)
 if hasattr(obj, '__code__')
 and obj.__code__ is code_obj][0]

现在您可以使用 inspect.signature()在过滤后的对象上。

---

<子>来自 gc.get_referrers(objs) 的免责声明:

_{This function will only locate those containers which support garbage collection; extension types which do refer to other objects but do not support garbage collection will not be found.}

---

完整代码示例:

import inspect
import gc

def ERROR_Printer_Inspection ( stream = sys.stderr ) :
    """
    called in try: except: <here>
    prints the last error-traceback in the given "stream"
    includes signature and function arguments if possible
    """
    stream.write('Traceback (most recent call last):\n')
    etype, value, _ = sys.exc_info() # get type and value for last line of output
    frame_infos = inspect.trace() # get frames for source-lines and arguments

    for f_idx, f_info in enumerate(frame_infos):
        stream.write(f'  File "{f_info.filename}", line {f_info.lineno}, in {f_info.function}\n')
        for line in f_info.code_context: # print location and code parts
            stream.write(f'    {line.lstrip()}')

        if f_idx+1 < len(frame_infos): # signature and arguments
            code_obj = frame_infos[f_idx+1].frame.f_code # codeobject from next frame
            function_obj = [obj for obj in gc.get_referrers(code_obj) if hasattr(obj, '__code__') and obj.__code__ is code_obj]

            if function_obj: # found some matching object
                function_obj=function_obj[0] # function_object
                func_name = frame_infos[f_idx + 1].function # name 
                stream.write(f'    > {func_name} {inspect.signature(function_obj)}\n')

            next_frame_locals = frame_infos[f_idx+1].frame.f_locals # calling arguments
            # filter them to the "calling"-arguments
            arguments = dict((key, next_frame_locals[key]) for key in code_obj.co_varnames if key in next_frame_locals.keys())
            stream.write(f'    -> {str(arguments)[1:-1]}\n')

    stream.write(f'{etype.__name__}: {value}\n')
    stream.flush()

---

问题:

如果在函数启动后对其进行编辑，则显示“调用”参数可能会产生误导:

def foo (a, b, **kwargs):
    del a, kwargs
    b = 'fail'
    return 0/0

try: foo(0, 1, test=True)
except: ERROR_Printer_Inspection()

输出:

Traceback (most recent call last):
  File "C:/test/errorHandler.py", line 142, in <module>
    try: foo(0, 1, test=True)
    > foo (a, b, **kwargs)
    -> 'b': 'fail'
  File "C:/test/errorHandler.py", line 140, in foo
    return 0 / 0
ZeroDivisionError: division by zero

你不能相信那个，但这是另一个问题的问题。

---

链接:

如果您想自己研究，这里有一些链接:

• Getting the Python function for a code object
• Python-Docs: inspect-module
• Python-Docs: Data model
• Python-Docs: garbage collector
• late.am - Exploring Python Code Objects