为什么不使用S :: x?

本文介绍了为什么不使用S :: x?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

从 cppreference 中考虑此示例:

struct S { static const int x = 1; }; void f() { &S::x; } // discarded-value expression does not odr-use S::x

我同意&S::x是一个舍弃值表达式，因为该标准指出(9.2，第 n4700 )

表达语句具有以下形式

expression-statement: expression_opt ;

该表达式是一个废弃值表达式(第8条)...

但是，足以让S::x不被 odr-使用吗? 6.2，第3段[basic.def.odr]声明

变量x的名称显示为可能评估的表达式ex，除非

使用 odr

• ...
• 如果x是对象，则 ex是表达式e 的潜在结果集中的元素，其中
  • 将左值到右值转换(7.1)应用于e或
  • e是一个舍弃值表达式(第8条).

问题是舍弃值表达式&S::x没有潜在结果(这意味着S::x不是&S::x的潜在结果)，如您从6.2第2段中所见[basic.def .odr]:

...表达式e的潜在结果集定义如下:

• 如果e是id表达式(8.1.4)，则该集合仅包含e.
• 如果e是具有数组操作数的下标操作(8.2.1)，则该集合包含该操作数的可能结果.
• ...
• 否则，该集合为空.

然后，您如何解释S::x不是未使用?

解决方案

它确实是使用过的.您的分析是正确的(我已经修正了前一阵子)

Consider this example from cppreference:

struct S { static const int x = 1; }; void f() { &S::x; } // discarded-value expression does not odr-use S::x

I agree that &S::x is a discarded-value expression, since the standard says (9.2, paragraph 1 [stmt.expr] from n4700)

Expression statements have the form

expression-statement: expression_opt ;

The expression is a discarded-value expression (Clause 8)...

However, is that enough for S::x to not be odr-used? 6.2, paragraph 3 [basic.def.odr] states

A variable x whose name appears as a potentially-evaluated expression ex is odr-used by ex unless

• ...
• 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 (7.1) is applied to e, or
  • e is a discarded-value expression (Clause 8).

The problem is that the discarded-value expression &S::x has no potential results (which means that S::x is not a potential result of &S::x), as you can see from 6.2, paragraph 2 [basic.def.odr]:

... The set of potential results of an expression e is defined as follows:

• If e is an id-expression (8.1.4), the set contains only e.
• If e is a subscripting operation (8.2.1) with an array operand, the set contains the potential results of that operand.
• ...
• Otherwise, the set is empty.

Then, how can you explain that S::x is not odr-used?

解决方案

It is indeed odr-used. Your analysis is correct (and I fixed that example a while ago).