python - 为什么在 __enter__ 引发异常时不执行 __exit__

Question

我最近想知道当 __enter__ 引发异常时不隐式调用 __exit__ 的原因是什么？为什么要这样设计？我正在实现服务运行器类以供 'with' 关键字使用，结果 __exit__ 从未被调用。

例子:

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.

和

Service1 running...
Service2 running...
Service3 failed!

函数 __exit__ 没有被调用。 Service1 和 Service2 没有被释放。

我可以将 _allocate_resource() 移动到 __init__ 但是类在这种用法中不是很有用:

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.
All services freed.

服务没有重新启动。

我可以重新实现 __enter__ 来处理异常，但它会向函数添加一些样板代码:

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？

Answer 1

如果您未能进入上下文，则没有理由尝试退出它，即如果您未能分配资源，则没有理由尝试释放它。

IIUC，您要找的只是:

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