关于类成员访问表达式的一个定义规则 [英] One definition rule about class member access expression

问题描述

在N4296,3.2 [basic.def.odr] p3：

变量 x ex 是由 ex - 到-rvalue转换为 x 产生一个不调用任何非平凡函数的常量表达式，如果 x 是一个对象， ex 是表达式 e 的一组潜在结果的一个元素，其中lvalue-对 e 应用到值转换，或者 e 是丢弃值表达式。




如何解释这一段？我发现了两个解释。

1从这里

<我们将这个分解为步骤：表达式中的变量`x`的出现构成一个odr使用，除非：

1. 或



2. 必须满足以下所有条件：
   
   
   
   1. 将左值到右值转换应用到 x 产生一个不调用任何非平凡函数的常量表达式 和
   
   
   2. ex 是表达式 e 的潜在结果集合中的一个元素， b>：
      
      
      
      1. 左值到右值转换应用于 e
      
      
      2. 或 e 是舍弃值
      
      
      
      
      
      

      和2来自cppreference http://en.cppreference.com/w/cpp/language/definition

      


      一个变量<$ c除非任何成立，否则在潜在求值表达式 ex 中的$ c> x

      
      
      

      
      

      • 将左值到右值转换应用到 x 不调用非平凡函数的表达式

      
      

      • x 是一个对象，ex是下列之一较大表达式的潜在结果 e ，其中较大的表达式是丢弃值表达式或左值到右值转换

      
      

      
      

      关于两个规则的第一个答案是和，另一个是强>。哪一个是正确的？

      
      
      

      请将规则分成多个步骤来说明此代码：

        struct S {static const int x = 0; }; 
       extern S s; //没有定义s 
       int i = s.x; //是s odr使用？是x odr使用？ 
       // gcc 5.1.0 is ok 
        

      
      

      解决方案

      cppreference 是是错误的;从标准中的语言（无论哪个版本）都清楚，两个子句必须保持。

      
      
      

      在您的示例中， s 不是常量表达式（C ++ 14：不满足出现在常量表达式中的要求），所以使用odr。

      
      
      

      同时， x 也使用odr，因为虽然可能在适当的上下文中的常量表达式中使用 x （例如作为 S 定义的数组） x 不是封闭表达式 sx 的潜在结果之一，这是唯一的封闭表达式丢弃值转换或左值到右值转换。

      
      
      

      gcc可能没有 s的定义或 x ，但没有要求实施诊断每个odr违例。

      
      

      In N4296, 3.2 [basic.def.odr]p3:

      A variable x whose name appears as a potentially-evaluated expression ex is odr-used by ex unless applying the lvalue-to-rvalue conversion to x yields a constant expression that does not invoke any non-trivial functions and, if x is an object, ex is an element of the set of potential results of an expression e, where either the lvalue-to-rvalue conversion is applied to e, or e is a discarded-value expression.

      How to explain this paragraph? I found two explanation.

      1 from here "Trying to understand [basic.def.odr]/2 in C++14 (N4140)"

      Let's split this into steps: The occurrence of a variable `x` in an expression `ex` constitutes an odr-use unless:

      1. Either ex is not potentially evaluated, or
      2. All of the following must be fulfilled:
         1. "applying the lvalue-to-rvalue conversion to x yields a constant expression that does not invoke any non-trivial functions" and
         2. "ex is an element of the set of potential results of an expression e" and either of the following holds:
            1. "either the lvalue-to-rvalue conversion is applied to e"
            2. "or e is a discarded-value expression"

      and 2 from cppreference http://en.cppreference.com/w/cpp/language/definition

      a variable x in a potentially-evaluated expression ex is odr-used unless any of the following is true:

      • applying lvalue-to-rvalue conversion to x yields a constant expression that doesn't invoke non-trivial functions

      • x is an object and ex is one of the potential results of a larger expression e, where that larger expression is either a discarded-value expression or an lvalue-to-rvalue conversion

      First answer about two rules is and, the other is any. Which one is right?

      Please split rules into steps to explain this code:

      struct S { static const int x = 0; };
      extern S s;// no definition of s
      int i = s.x;// is s odr-used? is x odr-used?
                  // gcc 5.1.0 is ok
      

      解决方案

      cppreference is was wrong; it is clear from the language in the standard (whichever version) that both subclauses must hold. I've corrected it.

      In your example, s is not a constant expression (C++14: does not satisfy the requirements for appearing in a constant expression) so is odr-used. The second subclause does not arise.

      Meanwhile, x is also odr-used, because although it would be possible to use x in a constant expression in an appropriate context (e.g. as an array bound within the definition of S); x is not one of the potential results of the enclosing expression s.x, which is the only enclosing expression subject to either the discarded-value transformation or the lvalue-to-rvalue conversion.

      gcc might be OK without a definition of s or x, but there's no requirement that an implementation diagnose every odr violation.

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