参考表达式的求值
根据@Potatoswatter 的建议，我创建了一个新的讨论。
参考是来自 @Potatoswatter 的 此响应
给出代码片段，
int i = 3, &j = i;
j = ++ i;
我寻求澄清的评论是这样的。 （这似乎是我对无序评估（又称序列点）的理解中一个重要的缺失部分）：
@Chubsdad：尽管它是一个别名， 它的左值评估不 需要对 i 进行左值评估。 一般来说，评估一个 参考文献不需要 手头上的原始对象。有 没有理由它应该是 UB，所以它使得 感觉应该有一个容易的漏洞 或转换为不是的代码 UB。
和
参考文献没有告诉 编译器去看看引用的 变量并获取其左值，因为 它可能不知道变量是什么 参考。编译器计算出 参考的左值和 左值标识一个对象。如果你 想进一步辩论这个问题，请 提出一个新问题。
问题中任何可能的不明确都是“未定义行为”的一部分，我正在尝试理解 C++0x 中的“无序求值”、“序列点”等。
As per @Potatoswatter's suggestion, I have created a new discussion.
Reference is this response from @Potatoswatter
Given the code snippet,
int i = 3, &j = i;
j = ++ i;
The comment which I seek clarity on, is this. (which seems to be an important missing piece in my understanding of the unsequenced evaluation a.k.a sequence point):
@Chubsdad: Even though it's an alias,
its glvalue evaluation does not
require a glvalue evaluation of i.
Generally speaking, evaluating a
reference does not require the
original object to be on hand. There's
no reason it should be UB, so it makes
sense there should be an easy loophole
or transformation to code which is not
UB.
and
The reference doesn't tell the
compiler to go look at the referenced
variable and get its lvalue, because
it might not know what variable is
referenced. The compiler computes the
lvalue of the reference and that
lvalue identifies an object. If you
want to debate this further, please
open a new question.
Any possible lack of clarity in the question is part of the 'undefined behavior' I am going through trying to understand 'unsequenced evaluation', 'sequence point' etc in C++0x.
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（2）
想象一下
如果您说
i是另一个名称的别名 - 什么名称？引用可以引用对象或函数。当您说“glvalue”时，您指的是特定表达式的属性，而不是对象的属性。现在，
i是一个左值表达式，ri也是一个左值表达式（都是语法类别id-expression）。它们命名（通过名称查找找到）一个引用和一个int变量。如果您现在确定 ri 情况的对象标识，则需要获取引用并使表达式引用其初始化所用的对象。这称为左值评估，因为您确定左值（即指示对象）的属性。
您需要对
i情况执行相同的操作。即，您找出左值表达式 i 引用的对象。因此，ri的泛左值评估与i的泛左值评估不同，尽管两者产生相同的结果。右值求值意味着获取左值并将左值应用到右值转换。换句话说，读取一个值。
Imagine the following
If you say that
iis an alias for another name - what name? A reference either references an object or function. When you say "glvalue", you refer to a property of a particular expression, not to a property of an object.Now,
iis an lvalue expression andriis an lvalue expression too (both of the syntactic categoryid-expression). They name (as found by name-lookup) a reference and anintvariable.If you now determine the object identity for the
ricase, you need to take the reference and make the expression refer to the object it was initialized with. This is called an lvalue evaluation because you determine the property of an lvalue (i.e the referent).You need to do the same for the
icase. I.e you figure out to what object the lvalue expressionirefers to. A glvalue evaluation ofrithus is different than glvalue evaluation ofi, despite both of them yielding the same result.Rvalue evaluation means to take an lvalue and apply the lvalue to rvalue conversion. In other words, to read a value.
从概念上讲，C++ 中的引用是某个对象的别名或替代名称。当涉及引用时，这个概念应该指导您解释语言规则。
本质上有两种实现 C++ 引用的方法：
在所提供的示例中，永远不可能存在未绑定到对象
i的引用j，因此编译器很可能会使用符号表实现参考注解方法。这意味着，在声明i和j后，它们可以在代码中互换使用，而不会对生成的代码或行为是否正确的问题产生任何影响。定义的。Conceptually, a reference in C++ is an alias, or alternative name, for some object. It is this concept that should guide you in the interpretation of the language rules when references are involved.
There are essentially two ways of implementing C++ references:
In the presented example, there is never a way there can be a reference
jthat is not bound to the objecti, so the compiler will most likely use the symbol-table annotation method of implementing the reference. This means that, after the declaration of bothiandj, they can be used interchangeably in the code without any effect on the generated code or on the question if the behaviour is defined.