Python中的静态变量？ [英] static variables in Python?

问题描述

又一个菜鸟问题...


有没有办法在Python中模仿C'的静态变量？或者是什么

喜欢它？我们的想法是为一个给定的函数配备一组仅属于它的

常量，以免混淆全局

名称空间中包含其他地方不需要的变量。 。


例如，在Perl中可以定义一个像这样的函数foo


* foo = do {

我的$ x = expensive_call（）;

sub {

返回do_stuff_with（$ x，@_）;

}

};


在这种情况下，foo是通过为其分配一个闭包来定义的，该闭包具有关联变量$ x，其中包含$ x范围。


Python中是否有等价物？


谢谢！


kynn

-

注意：在我的地址中，第一个时期之前的所有内容都是向后的;

和最后一个时期以及之后的所有内容都应该被丢弃。

解决方案

x = expensive_call（）;

sub {

返回do_stuff_with（

X， @_）;

}

};


在这种情况下，foo是通过为其分配一个闭包来定义的

一个关联的变量，

x，在它的范围内。


Python中是否有等价的？


谢谢！


kynn

-

注意：在我的地址中所有内容之前第一个时期是倒退;

和最后一个时期以及之后的所有时期都应该被丢弃。

Yet another noob question...

Is there a way to mimic C''s static variables in Python? Or something
like it? The idea is to equip a given function with a set of
constants that belong only to it, so as not to clutter the global
namespace with variables that are not needed elsewhere.

For example, in Perl one can define a function foo like this

*foo = do {
my $x = expensive_call();
sub {
return do_stuff_with( $x, @_ );
}
};

In this case, foo is defined by assigning to it a closure that has
an associated variable, $x, in its scope.

Is there an equivalent in Python?

Thanks!

kynn
--
NOTE: In my address everything before the first period is backwards;
and the last period, and everything after it, should be discarded.

解决方案

x = expensive_call();
sub {
return do_stuff_with(

x, @_ );
}
};

In this case, foo is defined by assigning to it a closure that has
an associated variable,

x, in its scope.

Is there an equivalent in Python?

Thanks!

kynn
--
NOTE: In my address everything before the first period is backwards;
and the last period, and everything after it, should be discarded.

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