Python - 当异常会被抛出时启动交互式调试器 [英] Python - start interactive debugger when exception would be otherwise thrown

问题描述

有没有办法让一个python程序启动一个交互式调试器，像什么 import pdb; pdb.set_trace（）而不是真的抛出一个异常？

我知道做这个工作的难度，但它会更有价值比一个巨大的堆栈跟踪，之后我必须使用来找出插入断点的位置，然后重新启动程序来进行调试。我知道简单地使调试器开始而不是抛出异常是没有意义的，因为任何异常都可以被捕获在一个级别或另一个级别，所以如果我可以选择一个异常的列表，交互式调试会话将会启动而不是被抛出（因为我知道这个列表中的异常真的是错误，之后没有有意义的程序行为）...

我听说Common Lisp有这样的东西，但我不知道它是如何正常工作的，只是这个真正的lispers赞美了很多...

解决方案

最简单的方法是将整个代码封装在 try 块中，如下所示：

 如果__name__ =='__main__'：
 
 try：
 raise异常（）
除了
 import pdb 
 pdb.set_trace（）
  

有一个更复杂的解决方案使用s sys.excepthook 来覆盖未捕获异常的处理，如这个配方

  ## {{{http://code.activestate.com/recipes/65287/（r5）
＃code snippet，要包含在'sitecustomize.py'
 import sys 
 
 def info（type，value，tb）：
如果hasattr（sys，'ps1'）或不是sys.stderr.isatty（）：
＃我们处于交互模式，或者我们没有一个类似tty的
＃设备，所以我们调用默认钩子
 sys .__ excepthook __（type，value，tb）
 else：
 import traceback，pdb 
＃我们不是在交互模式，打印例外... 
 traceback.print_exception（type，value，tb）
 print 
＃...然后在验尸模式下启动调试器。 
 pdb.pm（）
 
 sys.excepthook = info 
 ## http://code.activestate.com/recipes/65287/}}}} 
  

上述代码应该包含在sitecustomize.py中，这是由python自动导入的。调试器仅在python以非交互模式运行时启动。

Is there any way to make a python program start an interactive debugger, like what import pdb; pdb.set_trace() instead of actually throwing an exception?

I know the difficulty of making this work, but it would be much more valuable than a huge stack trace after which I have to use to figure out where to insert breakpoints and then restart the program to debug it. I know that simply making the debugger start instead of throwing an exception would not make sense because any exception can be caught at a level or another, so if I could just select a list of exceptions for which an interactive debug session would start instead of them being thrown (because I know the exceptions in this list would really be "errors" and no meaningful program behavior could follow afterwards)...

I've heard that Common Lisp has something like this, but I don't know how it works exactly, just that "true lispers" praise it a lot...

解决方案

The simplest way is to wrap your entire code inside a try block like this:

if __name__ == '__main__':

    try:
        raise Exception()
    except:
        import pdb
        pdb.set_trace()

There is a more complicated solution which uses sys.excepthook to override the handling of uncaught exceptions, as described in this recipe

## {{{ http://code.activestate.com/recipes/65287/ (r5)
# code snippet, to be included in 'sitecustomize.py'
import sys

def info(type, value, tb):
   if hasattr(sys, 'ps1') or not sys.stderr.isatty():
      # we are in interactive mode or we don't have a tty-like
      # device, so we call the default hook
      sys.__excepthook__(type, value, tb)
   else:
      import traceback, pdb
      # we are NOT in interactive mode, print the exception...
      traceback.print_exception(type, value, tb)
      print
      # ...then start the debugger in post-mortem mode.
      pdb.pm()

sys.excepthook = info
## end of http://code.activestate.com/recipes/65287/ }}}

The above code should be included in 'sitecustomize.py', which is automatically imported by python. The debugger is only started when python is run in non-interactive mode.

这篇关于Python - 当异常会被抛出时启动交互式调试器的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！