N3936第3.3.7 / 1节中的规则3是否冗余？ [英] Is rule 3 in section 3.3.7/1 from N3936 redundant?

问题描述

我最近回答了一个问题，涉及违反草案C ++ 14标准：N4140 节 3.3.7 > 段落 1 $ b $ b

I recently answered a question dealing with a violation of draft C++14 standard: N4140 section 3.3.7 Class scope paragraph 1 rule 2 which says:

在S类中使用的名称N在其
上下文中引用相同的声明，并在S的完成范围内重新求值。No $ b $违反此规则需要诊断。 $ 3

A name N used in a class S shall refer to the same declaration in its context and when re-evaluated in the completed scope of S. No diagnostic is required for a violation of this rule.

< c>也似乎相关，它说：

at the time rule 3 also seemed relevant and it says:

如果重新排序类中的成员声明产生一个备用的有效
程序在）和（2），程序是错误的，没有诊断是
需要。

If reordering member declarations in a class yields an alternate valid program under (1) and (2), the program is ill-formed, no diagnostic is required.

我的初步反应是，规则 3 似乎是多余的，只是对规则 2 的澄清，并不包括任何未被覆盖的案例。导致交替有效程序的重新排序也必须违反规则 2

My initial reaction is that rule 3 seems redundant and is really just a clarification of rule 2 and does not cover any cases not already covered. A reordering that results in a alternate valid program would also have to violate rule 2.

推荐答案

根据这些规则，到缺陷报告1875：在类作用域中重新排序声明规则 3 是多余的，建议的解决方案是删除规则 3 ，它说：

According to Defect Report 1875: Reordering declarations in class scope rule 3 is redundant and the proposed solution is to remove rule 3, it says:

规则＃3的需要不清楚;看起来任何
否则有效的重新排序将不得不违反规则＃2以便
产生不同的解释。从字面上来看，规则＃3也将$ b​​ $ b适用于简单地重新排序非静态数据成员，没有名称
依赖。是否可以简单删除？

The need for rule #3 is not clear; it would seem that any otherwise-valid reordering would have to violate rule #2 in order to yield a different interpretation. Taken literally, rule #3 would also apply to simply reordering nonstatic data members with no name dependencies at all. Can it be simply removed?

建议的解决方案是：

删除3.3.7 [basic.scope.class]段落1的第三项，
对后续项重新编号

Delete the third item of 3.3.7 [basic.scope.class] paragraph 1 and renumber the succeeding items

虽然这个缺陷报告似乎证实了我最初的怀疑，但我仍然有一种ging叨的感觉，或许规则 3 毕竟有点宽。 3.3.7 包括以下示例：

Although this defect report seems to confirm my initial suspicious I am left with a nagging feeling that perhaps rule 3 is bit broader after all. Section 3.3.7 includes the following example:

enum { i = 1 };

class X {
  char v[i]; // error: i refers to ::i
             // but when reevaluated is X::i
  int f() { return sizeof(c); } // OK: X::c
  char c;
  enum { i = 2 };
};

违反规则 2 c $ c> 3 但小的调整：

which violates both rule 2 and 3 but a small tweak:

enum { i = 1 };

class X {
  enum { i = 2 };
  char v[i];  // no longer refers to ::i 
              // but reordering can cause it to refer to ::i again

  int f() { return sizeof(c); } // OK: X::c
  char c;
};

似乎不再违反规则 2 肯定违反规则 3 。我认为这个代码示例是麻烦的，因为成员的重新排序很容易导致代码回溯违反规则2，但没有诊断需要指示这使得这个代码相当脆弱。

seems to no longer violate rule 2 but sure seems to violate rule 3. I would consider this code example to be troublesome since a reordering of the members could easily cause the code to be back in violation of rule 2 but no diagnostic is required to indicate this which makes this code rather fragile.

更新

据我理解规则 3 不适用于Casey在注释中提到的这个例子：

As far as I understand rule 3 does not apply to this example mentioned by Casey in the comment:

class X { int a; int b; };

，因为即使有多个有效排序，这种情况也不属于 1 和 2 其中规则 3 需要：

because even though there is more than one valid ordering this case does not fall under both rule 1 and 2 which rule 3 requires:

（1）和（2）下的备用有效程序

alternate valid program under (1) and (2)

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