从FrameInfo或python中的frame获取函数签名

时间:2018-10-09 07:23:34

标签: python traceback

在用python编写我自己的异常钩子的同时,我想到了使用inspect-模块为自己提供有关如何调用该函数的更多信息的想法。

这意味着函数的签名以及传递给它的自变量

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')

在像这样的基本示例中,这可以很好地工作:

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,一旦您留下1个文件,就无法将函数名称映射到基本名称空间。

  

文件1:import file2; try: file2.foo() except: ...
  文件2:import file3; def foo(): file3.foo()
  文件3:def foo(): return 0/0

非常重要,我正在寻找一种从<function foo at 0x000002F4A43ACD08>FrameInfo对象获取函数(如frame)的方法,但是我看到的唯一信息是名称以及文件和行。
(我不喜欢通过在特定行中查看源文件来获得签名的想法。)

到目前为止,最佳参考文献是Inspect-documentation,但我还没有发现有用的东西。

1 个答案:

答案 0 :(得分:0)

基于this answerjsbueno,我找到了一种用于恢复签名的解决方案。

使用gc(垃圾收集器)功能get_referrers(),您可以搜索所有直接引用特定对象的对象。

使用框架f_code提供的代码对象,您可以使用此功能查找框架以及该功能。

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]

现在您可以在过滤的对象上使用inspect.signature()


来自gc.get_referrers(objs)的免责声明:

  

此函数将仅找到那些支持垃圾收集的容器;找不到引用其他对象但不支持垃圾回收的扩展类型。   


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”参数可能会误导用户:

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

您不能相信,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: