Python 为什么uuuu enter uuuu引发异常时不执行uuuu exit uuuuuuuu

我最近想知道当

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu>引发异常时，没有隐式调用
的原因是什么？
为什么它是这样设计的？我实现了一个ServiceRunner类，该类可以通过
'with'
关键字使用，结果证明，
\uuuuuuuuuuuu
从未被调用

例如：

class ServiceRunner(object):
    def __init__(self, allocation_success):
        self.allocation_success = allocation_success

    def _allocate_resource(self):
        print("Service1 running...")
        print("Service2 running...")

        # running service3 fails ...
        if not self.allocation_success:
            raise RuntimeError("Service3 failed!")
        print("Service3 running...")

    def _free_resource(self):
        print("All services freed.")

    def __enter__(self):
        self._allocate_resource()
        return self

    def __exit__(self, exc_type, exc_val, exc_tb):
        self._free_resource()

用法：

with ServiceRunner(allocation_success=True):
    pass

try:
    with ServiceRunner(allocation_success=False):
        pass
    except Exception as e:
        print(e)

输出：

Service1 running...
Service2 running...    
Service3 running...
All services freed.

Service1 running...
Service2 running...
Service3 running...
All services freed.
All services freed.

及

未调用函数
\uuuuu exit\uuuu
。服务1和服务2未被释放

我可以将
\u allocate\u resource（）
移动到
\uuu init\uuu
，但在这种情况下，类不是很有用：

try:
    runner = ServiceRunner(allocation_success=True)
except Exception as e:
    print(e)
else:
    with runner as r:
        r.do()

    with runner as r:
        r.do()

输出：

Service1 running...
Service2 running...    
Service3 running...
All services freed.

Service1 running...
Service2 running...
Service3 running...
All services freed.
All services freed.

服务不会再次启动

我可以重新实现
\uuuu enter\uuuu
来处理异常，但它为函数添加了一些样板代码：

def __enter__(self):
    try:
        self._allocate_resource()
    except Exception as e:
        self.__exit__(*sys.exc_info())
        raise e

这是最好的解决方案吗？
如果您未能输入上下文，则没有理由尝试退出它，也就是说，如果您未能分配资源，则没有理由尝试释放它

IIUC，你要找的只是：

try:
   with ServiceRunner() as runner:
       runner.do()
except Exception as e:
    print(e)

为什么默认情况下会调用
\uuuuuuuuuuuuuuuuuuuuu
？如果我将open（'file'u that'u not'u存在'）作为infle:
，会发生什么？因为文件不存在，所以没有使用的资源被关闭。@roganjosh:那是另一种情况。如果该文件不存在，则永远不会调用
\uuu\uu enter\uu
。在引发异常之前，已分配了一些资源。我想确定他们被释放了。那么你在
\uuuu\uuu\uuu
上尝试/捕获的方法就可以了；不要觉得有义务使用
\uuuuu退出\uuuu