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

Question

如果发生异常，使用'with-open'时，是否需要finally块来确保文件关闭。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，假设它确实打开了文件，它会被关闭还是我需要关闭它？同样，如果yaml.load((抛出，那么with open是否仍然完成，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()

但现在a_file不在范围内，对吧？

Answer 1

with open(filepath, mode) as a_file语句只在该段内部保持文件打开，当它"退出"；无论在open()或yaml.load(...)中的哪里抛出异常，都不需要a_file.close()。