多项任务评估辩论 [英] Multiple Assignment Evaluation Debate

问题描述

大家好，


一位同事和我一直在争论

对a = b = c这句话的正确预期。两个不同版本的GCC

最终将其编译为b = c; a = b而另一个结束了

将其编译为a = c; b = c。 （两者均未启用优化）。

我们如何注意到您可能会问？好吧，在我们的例子中''b''是一个内存

映射寄存器，它只有一个可选择的可写位数。他/ b $ b声称它是一个'C标准'，它将被评估为a = c; b

= c。我个人认为将b = c评估为b $ b更有意义。 a = b，虽然我永远不会编写具有

可疑操作的代码。任何人都可以解决这个争论吗？

Hi All,

A coworker and I have been debating the ''correct'' expectation of
evaluation for the phrase a = b = c. Two different versions of GCC
ended up compiling this as b = c; a = b and the other ended up
compiling it as a = c; b = c. (Both with no optimizations enabled).
How did we notice you may ask? Well, in our case ''b'' was a memory
mapped register that only has a select number of writable bits. He
claims it has been a ''C Standard'' that it will be evaluted as a = c; b
= c. I personally believe that it would make more sense for it to be
evaluated as b = c; a = b, although I would never write code that has a
questionable operation. Can anyone settle this debate?

推荐答案

be * **@uwalumni.com 说：

大家好，

一位同事和我一直在争论'正确'的期望
评估短语a = b = c。两个不同版本的GCC
最终将其编译为b = c; a = b而另一个最终将其编译为a = c; b = c。 （两者均未启用优化）。
我们如何注意到您可能会问？好吧，在我们的例子中，''b''是一个内存
映射寄存器，只有一个可选择的可写位数。他声称它是一个'C标准'，它将被评估为a = c; b
= c。我个人认为将它评估为b = c会更有意义。 a = b，虽然我永远不会编写具有可疑操作的代码。任何人都可以解决这个争论吗？

Hi All,

A coworker and I have been debating the ''correct'' expectation of
evaluation for the phrase a = b = c. Two different versions of GCC
ended up compiling this as b = c; a = b and the other ended up
compiling it as a = c; b = c. (Both with no optimizations enabled).
How did we notice you may ask? Well, in our case ''b'' was a memory
mapped register that only has a select number of writable bits. He
claims it has been a ''C Standard'' that it will be evaluted as a = c; b
= c. I personally believe that it would make more sense for it to be
evaluated as b = c; a = b, although I would never write code that has a
questionable operation. Can anyone settle this debate?

结合性是：a =（b = c）;但a，b和

c的评估顺序未指定。标准只要求a和b

的新值在前一个时间之后的某个时间提供，并且在下一个之后，

序列点。


因此，gcc完全有权将其编译为：a = c; b = c;或b = c;

a = c;或b = c; a = b;或tmp = c; a = tmp; b = tmp;或任何其他

组合给出相同的结果。


如果你依赖于评估的顺序，a，b和c的任何副作用< b / b
最终可能会咬你。


-

Richard Heathfield

" Usenet is一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk

电子邮件：rjh在上面的域名（但显然放弃了www）

The associativity is: a = (b = c); but the order of evaluation of a, b, and
c is unspecified. The Standard only requires that the new values of a and b
are provided at some time after the previous, and before the following,
sequence point.

So gcc is quite within its rights to compile it to: a = c; b = c; or b = c;
a = c; or b = c; a = b; or tmp = c; a = tmp; b = tmp; or any other
combination that gives the same result.

If you depend on the order of evaluation, any side effects of a, b, and c
are likely to bite you eventually.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

在2006-03-10， be *** @ uwalumni.com < be *** @ uwalumni.com>写道：

On 2006-03-10, be***@uwalumni.com <be***@uwalumni.com> wrote:

大家好，

一位同事和我一直在争论评估a = b = c这句话的'正确'期望。两个不同版本的GCC
最终将其编译为b = c; a = b而另一个最终将其编译为a = c; b = c。 （两者均未启用优化）。
我们如何注意到您可能会问？好吧，在我们的例子中，''b''是一个内存
映射寄存器，只有一个可选择的可写位数。


如果你那样做，那就是挥发性。关键字可以做一些事情

的可预测性

他声称它是一个'C标准'，它将被评估为a =
c; b = c。


他对标准的解释与我的相同。

我个人认为将其评估为更有意义b = c; a = b，虽然我永远不会编写具有可疑操作的代码。谁能解决这个争论？

Hi All,

A coworker and I have been debating the ''correct'' expectation of
evaluation for the phrase a = b = c. Two different versions of GCC
ended up compiling this as b = c; a = b and the other ended up
compiling it as a = c; b = c. (Both with no optimizations enabled).
How did we notice you may ask? Well, in our case ''b'' was a memory
mapped register that only has a select number of writable bits.
If you do that, a sprinkling of the "volatile" keyword may do something
for predictability
He claims it has been a ''C Standard'' that it will be evaluted as a =
c; b = c.
His interpretaion of the standard is the same as mine.
I personally believe that it would make more sense for it to be
evaluated as b = c; a = b, although I would never write code that has a
questionable operation. Can anyone settle this debate?

使用volatile。你可能仍然会得到错误的答案，但它更可能是一致的

。

Use volatile. You might still get the wrong answer, but it''s more likely
to be consistent.

Richard Heathfield写道：

Richard Heathfield wrote:

be***@uwalumni.com 说：

大家好，

一位同事和我一直在讨论对a = b = c这个短语进行评估的正确期望。两个不同版本的GCC
最终将其编译为b = c; a = b而另一个最终将其编译为a = c; b = c。 （两者均未启用优化）。
我们如何注意到您可能会问？好吧，在我们的例子中，''b''是一个内存
映射寄存器，只有一个可选择的可写位数。他声称它是一个'C标准'，它将被评估为a = c; b
= c。我个人认为将它评估为b = c会更有意义。 a = b，虽然我永远不会编写具有可疑操作的代码。谁能解决这个争论？

Hi All,

A coworker and I have been debating the ''correct'' expectation of
evaluation for the phrase a = b = c. Two different versions of GCC
ended up compiling this as b = c; a = b and the other ended up
compiling it as a = c; b = c. (Both with no optimizations enabled).
How did we notice you may ask? Well, in our case ''b'' was a memory
mapped register that only has a select number of writable bits. He
claims it has been a ''C Standard'' that it will be evaluted as a = c; b
= c. I personally believe that it would make more sense for it to be
evaluated as b = c; a = b, although I would never write code that has a
questionable operation. Can anyone settle this debate?

结合性是：a =（b = c）;但是a，b和
c的评估顺序是未指定的。标准只要求a和b
的新值在前一个之后，之后的序列点之前的某个时间提供。

所以gcc是相当的它有权将其编译为：a = c; b = c;或b = c;
a = c;或b = c; a = b;或tmp = c; a = tmp; b = tmp;或任何其他
组合产生相同的结果。

如果你依赖于评估的顺序，a，b和c
的任何副作用可能会咬你最终。

The associativity is: a = (b = c); but the order of evaluation of a, b, and
c is unspecified. The Standard only requires that the new values of a and b
are provided at some time after the previous, and before the following,
sequence point.

So gcc is quite within its rights to compile it to: a = c; b = c; or b = c;
a = c; or b = c; a = b; or tmp = c; a = tmp; b = tmp; or any other
combination that gives the same result.

If you depend on the order of evaluation, any side effects of a, b, and c
are likely to bite you eventually.

你错过了他的问题。无论与评估订单有什么关系，b的中间分配都可以影响分配给a的值。 int = unsigned char = int，

例如。如果b是易失性的，也可能会改变图片，

正如另一个线程所讨论的那样。假设a = b = c可以被b = c，a = b
$ b $取代，这是不安全的（因为

可以辩论） b即使b是挥发性的。但是说a = b = c意味着（无论是什么

顺序）a = c和b = c是完全错误的。

You missed the point of his question. Regardless of anything having
to do with evaluation order, the intermediate assignment to b can
affect the value that is assigned to a. int = unsigned char = int,
for example. If b is volatile that may also change the picture,
as another thread discusses. It isn''t safe to assume (because
it''s open to debate) that a = b = c may be replaced by b = c, a = b
even if b is volatile. But saying a = b = c means (in whatever
order) a = c and b = c is simply wrong.

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