a | = b，a | b和a || b（为什么不是|| = b？） [英] a |=b, a | b and a || b (why not a ||= b?)

问题描述

为什么不能使用|| = b？为了提高效率，我只想对
进行必要的评估。我必须把它写成更长的吗？方式（a =

a || b）？


谢谢，


Mats-Lennart

解决方案

Mats-Lennart Hansson< ap ******** @ hotmail.com>写道：

为什么不能使用|| = b？为了提高效率，我只想尽可能多地评估。我必须把它写成更长的吗？方式（a =
a || b）？

我怀疑它是因为那将是一个很少使用的运算符 - 它是'

我认为通常不会以这种方式使用布尔值。写作

在我看来，漫长的道路也更清晰。


-

Jon Skeet - < sk * **@pobox.com>
http://www.pobox.com/ 〜双向飞碟

如果回复小组，请不要给我发邮件

| =运算符就是你的''正在寻找。它适用于布尔值

值。


" Mats-Lennart Hansson"写道：

为什么不能使用|| = b？为了提高效率，我只想尽可能多地评估。我必须把它写成更长的吗？方式（a =
a || b）？

谢谢，

Mats-Lennart

Jeremy Davis< Je ********* @ discussion.microsoft.com>写道：

| =运算符是你正在寻找的。它在boolean
值上运行得很好。

是的，但它没有短路，这就是OP想要的，我是/>
相信。实际上：


if（！a）

{

a = b;

}


（其中b可能是方法调用或属性）。


-

Jon Skeet - < sk ***@pobox.com>
http://www.pobox.com /〜双向飞碟

如果回复小组，请不要给我发邮件

Hi,
Why isn''t it possible to use a ||= b? To increase efficiency, I only want to
evaluate as much as neccessary. Do I have to write it the "longer" way (a =
a || b)?

Thanks,

Mats-Lennart

解决方案

Mats-Lennart Hansson <ap********@hotmail.com> wrote:

Why isn''t it possible to use a ||= b? To increase efficiency, I only want to
evaluate as much as neccessary. Do I have to write it the "longer" way (a =
a || b)?

I suspect it''s because that would be a very rarely used operator - it''s
not that often that booleans are used in that way, I believe. Writing
it the long way is also clearer, in my view.

--
Jon Skeet - <sk***@pobox.com>
http://www.pobox.com/~skeet
If replying to the group, please do not mail me too

The |= operator is what you''re looking for. It works just fine on boolean
values.

"Mats-Lennart Hansson" wrote:

Hi,
Why isn''t it possible to use a ||= b? To increase efficiency, I only want to
evaluate as much as neccessary. Do I have to write it the "longer" way (a =
a || b)?

Thanks,

Mats-Lennart

Jeremy Davis <Je*********@discussions.microsoft.com> wrote:

The |= operator is what you''re looking for. It works just fine on boolean
values.

Yes, but it doesn''t short-circuit, which is what the OP wanted, I
believe. Effectively:

if (!a)
{
a = b;
}

(where b may be a method call or property).

--
Jon Skeet - <sk***@pobox.com>
http://www.pobox.com/~skeet
If replying to the group, please do not mail me too

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