不清楚类变量 [英] Unclear On Class Variables
问题描述
我有点困惑。我的印象是：
class foo（对象）：
x = 0
y = 1
表示x和y是一个类的所有实例共享的变量。
但是当我针对两个foo实例运行它时，设置值
$ b $ x和x，它们确实是*实例*的独特之处，而不是
类。
已经很晚了，我可能会失踪显而易见的。启蒙赞赏...
-
--------------------------- -------------------------------------------------
Tim Daneliuk tu****@tundraware.com
PGP Key ： http://www.tundraware.com/PGP/
Tim Daneliuk写道：
我有点困惑。我的印象是：
类foo（对象）：
x = 0
y = 1
表示x和y是变量由类的所有实例共享。
但是当我针对foo的两个实例运行它，并设置x和y的值
时，它们确实是* instance *的唯一而不是
上课。
现在已经很晚了，我可能会错过这个显而易见的事。启蒙
赞赏...
如果没有实际代码如何设置变量，没有人可以回答这个问题。
-
问候，
Diez B. Roggisch
2005年1月13日美国东部时间07:18:26，Tim Daneliuk< tu **** @ tundraware.com>写道：我有点困惑。我的印象是：
类foo（对象）：
x = 0
y = 1
表示x和y是变量由类的所有实例共享。
但是当我针对foo的两个实例运行它，并设置x和y的值
时，它们确实是* instance *的唯一而不是
class。
我明白为什么你会这么想：
< blockquote class =post_quotes> class Spam（object）：
.... eggs = 4
.... spam = Spam（）
spam2 = Spam（）
spam.eggs
4 spam2.eggs
4 spam.eggs = 2
spam.eggs
2 spam2.eggs
4
但是由于整数是不可变的，你被误导了。
''spam.eggs = 2''是*创建*实例成员 - 没有一个
之前。看看可变对象会发生什么：
类垃圾邮件（对象）：
.... eggs = [3]
.... spam =垃圾邮件（）
spam2 =垃圾邮件（）
spam.eggs
[3] spam2.eggs
[3] spam.eggs.append（5）
spam.eggs
[3,5] spam2.eggs
[3,5]
-
干杯，
Simon B，
si *** @ brunningonline.net ，
http：// www。 brunningonline.net/simon/blog/
你好蒂姆，
如果你有
< br $>
类Foo（对象）：
x = 0
y = 1
foo = Foo（）
foo.x＃读取实例或类属性（本例中为类）
foo.x = val＃设置实例属性（因为foo实例不是
类）
Foo.x = val＃设置一个类属性
foo .__ class .__ x = val＃做同样的事情
这可能有时令人困惑。恕我直言，以下特别是
讨厌：
foo = Foo（）
foo.x + = 1
打印foo.x
1打印Foo.x
0
虽然+ =运算符看起来像是在原地添加它不是。
它只是foo.x = foo.x + 1的语法糖。
- harold -
2005年1月13日07:18，Tim Daneliuk写道：
我有点困惑。我的印象是：
类foo（对象）：
x = 0
y = 1
表示x和y是变量由类的所有实例共享。
但是当我针对foo的两个实例运行它，并设置x和y的值
时，它们确实是* instance *的唯一而不是
上课。
现在已经很晚了，我可能会错过这个显而易见的事。启蒙
赞赏......
-
------------------------------- ----------------------------------------
----- < Tim Daneliuk tu****@tundraware.com
PGP密钥： http://www.tundraware.com/PGP/
-
http://mail.python.org/mailman/listinfo/python-list
-
每个人都是天才。
只是有些人实现它太愚蠢了。
I am a bit confused. I was under the impression that:
class foo(object):
x = 0
y = 1
means that x and y are variables shared by all instances of a class.
But when I run this against two instances of foo, and set the values
of x and y, they are indeed unique to the *instance* rather than the
class.
It is late and I am probably missing the obvious. Enlightenment appreciated ...
--
----------------------------------------------------------------------------
Tim Daneliuk tu****@tundraware.com
PGP Key: http://www.tundraware.com/PGP/
Tim Daneliuk wrote:
I am a bit confused. I was under the impression that:
class foo(object):
x = 0
y = 1
means that x and y are variables shared by all instances of a class.
But when I run this against two instances of foo, and set the values
of x and y, they are indeed unique to the *instance* rather than the
class.
It is late and I am probably missing the obvious. Enlightenment
appreciated ...
Without actual code how you set the vars, no one can answer that question.
--
Regards,
Diez B. Roggisch
On 13 Jan 2005 07:18:26 EST, Tim Daneliuk <tu****@tundraware.com> wrote:I am a bit confused. I was under the impression that:
class foo(object):
x = 0
y = 1
means that x and y are variables shared by all instances of a class.
But when I run this against two instances of foo, and set the values
of x and y, they are indeed unique to the *instance* rather than the
class.
I can see why you might think that:
class Spam(object): .... eggs = 4
.... spam = Spam()
spam2 = Spam()
spam.eggs 4 spam2.eggs 4 spam.eggs = 2
spam.eggs 2 spam2.eggs 4
But you are being mislead by the fact that integers are immutable.
''spam.eggs = 2'' is *creating* an instance member - there wasn''t one
before. Have a look at what happens with a mutable object:
class Spam(object): .... eggs = [3]
.... spam = Spam()
spam2 = Spam()
spam.eggs [3] spam2.eggs [3] spam.eggs.append(5)
spam.eggs [3, 5] spam2.eggs
[3, 5]
--
Cheers,
Simon B,
si***@brunningonline.net,
http://www.brunningonline.net/simon/blog/
Hi Tim,
If you have
class Foo(object) :
x = 0
y = 1
foo = Foo()
foo.x # reads either instance or class attribute (class in this case)
foo.x = val # sets an instance attribute (because foo is instance not
class)
Foo.x = val # sets a class attribute
foo.__class.__x = val # does the same
this might be sometimes confusing. IMHO, the following is especially
nasty:
foo = Foo()
foo.x += 1
print foo.x 1 print Foo.x0
although the += operator looks like an inplace add it isn''t.
it is just syntactic sugar for foo.x = foo.x + 1.
- harold -
On 13.01.2005, at 07:18, Tim Daneliuk wrote:
I am a bit confused. I was under the impression that:
class foo(object):
x = 0
y = 1
means that x and y are variables shared by all instances of a class.
But when I run this against two instances of foo, and set the values
of x and y, they are indeed unique to the *instance* rather than the
class.
It is late and I am probably missing the obvious. Enlightenment
appreciated ...
--
-----------------------------------------------------------------------
-----
Tim Daneliuk tu****@tundraware.com
PGP Key: http://www.tundraware.com/PGP/
--
http://mail.python.org/mailman/listinfo/python-list
--
Everyone is a genius.
It''s just that some people are too stupid to realize it.
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