为什么当__enter__引发异常时不执行__exit__

Question

最近我想知道当__exit__引发异常时不隐式调用__enter__的背后原因是什么？ 为什么这样设计？我正在实施服务运行程序类，以便可以通过'with'关键字使用，事实证明从未调用过__exit__。

示例：

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!

未调用函数__exit__。 Service1和Service2不会释放。

我可以将_allocate_resource()移到__init__，但是在这种用法中，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.
All services freed.

服务不会再次启动。

我可以重新实现__enter__来处理异常，但是它向该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？

Answer 1

如果您未能输入上下文，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，您正在寻找的只是：

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