如何隐藏回溯的最后一次调用？

Question

我有以下代码

def _raise_exception(exception):
    raise exception

def require_odd(n):
    if n % 2 == 0:
        _raise_exception(ValueError("n must be odd"))

print(require_odd(2))

当我运行它时，它会显示

Traceback (most recent call last):
  File "main.py", line 31, in <module>
    print(require_odd(2))
  File "main.py", line 29, in require_odd
    _raise_exception(ValueError("n must be odd"))
  File "main.py", line 25, in _raise_exception
    raise exception
ValueError: n must be odd

由于回溯的最后一行代码是多余的，我该如何引发异常以使其显示以下内容？

Traceback (most recent call last):
  File "main.py", line 31, in <module>
    print(require_odd(2))
ValueError: n must be odd

---

目前，我正在这样做：

def _raise_exception(exception):
    try:
        raise exception
    except Exception as e:
        # Remove the last two calls (from this function and from the caller of this function)
        stack_lines = traceback.format_stack()[:-2]

        # Get the exception lines that contains the "Traceback (most recent call last):" in
        # the first line and the "ValueError: n must be odd" in the last line
        exception_lines = traceback.format_exception(e.__class__, e, e.__traceback__)

        # Join the relevant lines, print to stderr and exit
        print(''.join(exception_lines[:1] + stack_lines + exception_lines[-1:]), file=sys.stderr)
        exit(1)

def require_odd(n):
    if n % 2 == 0:
        _raise_exception(ValueError("n must be odd"))

print(require_odd(2))

确实可以打印

Traceback (most recent call last):
  File "main.py", line 31, in <module>
    print(require_odd(2))
ValueError: n must be odd

但它只是感觉临时。有没有更好的办法？

Answer 1

你不能去掉_raise_exception 函数吗？

如果你只是跑

if n % 2 == 0: 
    raise ValueError("n must be odd") 

这将消除最后一行，以及该过程中不必要的步骤。它不会消除中间线，但一般来说，最佳实践（我知道）是不理会堆栈。当出现问题时，您需要尽可能多的数据。

如果我遗漏了什么，请告诉我，我会更新。