为什么不使用S :: x? [英] Why is S::x not odr-used?

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).

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