python - 对自定义 Exception 的 try/except 感到困惑

Question

我的代码:

class AError(Exception):
    print 'error occur'
for i in range(3):
    try:
        print '---oo'
        raise AError
    except AError:
        print 'get AError'
    else:
        print 'going on'
    finally:
        print 'finally'

当我运行上面的代码时，输​​出是这样的:

error occur
---oo
get AError
finally
---oo
get AError
finally
---oo
get AError
finally

我认为字符串"error occur"应该出现三次，比如"---oo"，但它只出现了一次；为什么？

Answer 1

澄清Paul's answer ，这是一个简单的例子:

class Test(object):

    print "Class being defined"

    def __init__(self):
        print "Instance being created"


for _ in range(3):
    t = Test()

此输出将是:

Class being defined
Instance being created
Instance being created
Instance being created

在 class 定义内但在方法 definitition 之外的代码 仅在定义类时运行一次。

如果您希望代码每当创建一个实例时运行，它应该在__init__ 实例方法(或者，偶尔， __new__ 类方法)。但是，请注意，如果您为子类定义 __init__，您应该确保它也调用父类(super class)的 __init__:

class AError(Exception):

    def __init__(self, *args, **kwargs):
        Exception.__init__(self, *args, **kwargs) # call the superclass
        print 'error occur' # print your message

这确保子类支持父类(super class)的参数；在 Exception 的情况下，您可以例如传递错误信息:

>>> raise AError("Something went wrong.")
error occur # your message gets printed when the instance is created

Traceback (most recent call last):
  File "<pyshell#11>", line 1, in <module>
    raise AError("Something went wrong.")
AError: Something went wrong. # the error message passes through to the traceback

有关 *args, **kwargs 语法的解释，如果您不熟悉它，请参阅例如What does ** (double star) and * (star) do for parameters? .您还可以使用 super 调用父类(super class)方法，参见例如Understanding Python super() with __init__() methods .