Python中的Doctest和Decorators [英] Doctest and Decorators in Python

问题描述

我试图使用Python装饰器来捕获异常并记录异常.

I was trying to use Python decorator to catch exceptions and log the exceptions.

import os.path
import shutil

class log(object):
    def __init__(self, f):
        print "Inside __init__()"
        self.f = f

    def __call__(self, *args):
        print "Inside __call__()"
        try:
            self.f(*args)
        except Exception:
            print "Sorry"

@log
def testit(a, b, c):
    print a,b,c
    raise RuntimeError()

if __name__ == "__main__":
    testit(1,2,3)

工作正常

Desktop> python deco.py 
Inside __init__()
Inside __call__()
1 2 3
Sorry

问题是，当我尝试用于doctest进行测试

The issue is that when I tried to use for testing with doctest

@log
def testit(a, b, c):
    """
    >>> testit(1,2,3)
    """
    print a,b,c
    raise RuntimeError()

if __name__ == "__main__":
    import doctest
    doctest.testmod()

似乎什么都没发生.

Desktop> python deco2.py 
Inside __init__()

这是怎么了?

推荐答案

修饰后的函数(实际上是一个类实例)没有获得原始函数的 __ doc __ 属性(即 doctest 进行解析).您可以将 __ doc __ 复制到类实例中，但是...老实说，我真的没有真正在这里看到类的需要-您会仅使用 functools.wraps

The decorated function (which is actually a class instance) doesn't get the __doc__ attribute of the original function (which is what doctest parses). You could just copy __doc__ over to the class instance, but ... Honestly, I don't really see the need for a class at all here -- you'd probably be better just using functools.wraps

import functools
def log(func):
    @functools.wraps(func)
    def wrapper(*args):
        try:
            return func(*args)
        except Exception:
            print "sorry"
    return wrapper

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