在 python (2.7) 中，当使用 'with open' 时，我是否需要一个 'finally close' 块以防抛出错误（参见示例）

Question

使用'with open'时，如果发生异常，我是否需要finally 块来确保文件已关闭。例如

try:
    with open(self.some_path, "r") as a_file:
        self.yaml_contents = yaml.load(a_file.read())
except IOError as ex:
    logger.error("Failed to open (or load) settings file '{}' because '{}'".format(self.some_path,ex.strerror))
    raise

如果 open() 抛出异常，假设它确实打开了文件，它会被关闭还是我需要关闭它？另外，如果 yaml.load() 抛出，那么 with open 是否仍会完成，即 close() 文件？

try:
    with open(self.some_path, "r") as a_file:
        self.yaml_contents = yaml.load(a_file.read())
except IOError as ex:
    logger.error("Failed to open (or load) settings file {} because {}".format(self.some_path,ex.strerror))
    raise
finally:
    a_file.close()

但是现在 a_file 不在范围内，对吗？

Answer 1

with open(filepath, mode) as a_file 语句使文件仅在该段内打开 - 当它“退出”该段时，不需要 a_file.close()，无论在何处抛出异常 - 在 open() 或 yaml.load(...).

在 python (2.7) 中，当使用 'with open' 时，我是否需要一个 'finally close' 块以防抛出错误（参见示例）

In python (2.7) when using 'with open', do I need a 'finally close' block in case of error thrown (see example)

python

python-2.7

try-catch

with-statement

try-catch-finally