如何避免重复异常处理？

Question

我已经将IOError处理移动到一个单独的函数,以避免在打开文件进行读取时出现样板.

但是如果在读取文件时IOError会触发怎么办？如果sshfs断开连接,或者文件被root删除等？

def safe_open(*args):
    try:
        return open(*args)
    except IOError:
        quit('Error when opening file \'{0}\'. Error #{1[0]}: {1[1]}'.format(\
        args[0], sys.exc_info()[1].args))

...

with safe_open(myfile, 'r') as f:
    for i in f:
        print i

with safe_open(anotherfile, 'r') as f:
    try:
        conf = ''.join(f).format(**args)
    except KeyError:
        quit('\nOops, your template \'{0}\' has placeholders for' + \
        'parameters\nthat were not supplied in the command line: - {1}\n' +
        '\nCan\'t proceed. Ending. Nothing has been changed yet.'.format( \
        args['host_template'], '\n - '.join(sys.exc_info()[1].args)), 1)

文件以不同的方式读取,所以我没有看到将它放入函数并将更改的部分作为参数传递的方法.

[补充:想到这个解决方案,但它使一个无法关闭的发电机.如果循环停止,则文件保持打开状态.]

def reader(*args):
    try:
        with safe_open(*args) as f:
            for i in f:
                yield i
    except IOError:
        print('IOError when trying to read \'{0}\''.format(args[0]))

for i in reader(myfile, 'r'):
    pass # do some job

Answer 1

我可能会尝试将文件操作移动到单独的函数中并将它们包装在try ...中除外

实际上我只是有一个更好的想法...将错误处理放入装饰器并将装饰器应用于执行文件操作的每个函数

def catch_io_errors(fn):
    def decorator(*args, **kwargs):
        try:
            return fn(*args, **kwargs)
        except IOError:
            quit('whatever error text')
    return decorator

然后你可以把所有文件操作放到他们自己的函数中并应用装饰器

@catch_io_errors
def read_file():
    with open(myfile, 'r') as f:
        for i in f:
            print i

或者如果你需要兼容python 2.3:

def read_file():
    f = open(myfile, 'r')
    for i in f:
        print i
    f.close()

read_file = catch_io_errors(read_file)