C++14 草案 (N3936) 在 §3.2/3 中声明:

A variable x whose name appears as a potentially-evaluated expression ex is odr-used unless applying the lvalue-to-rvalue conversion (4.1) to x yields a constant expression (5.19) 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 (4.1) is applied to e, or e is a discarded-value expression (Clause 5).

这对我来说没有任何意义:如果表达式 e 是 discarded-value 表达式 取决于上下文，其中 e。 expression-statement(第 6.2 节)中使用的每个表达式都是 discarded-value 表达式。 左值到右值的转换是否应用于e也取决于e所使用的上下文。

此外，一个表达式出现在另一个表达式的潜在结果集合中意味着什么。一个人需要一个表达式相等的概念才能确定一个集合的成员。但我们没有参照透明性，所以我看不出这是如何实现的。

为什么从 C++11 改为 C++14？这应该如何解释？就目前而言，这没有任何意义。

最佳答案

odr使用的目的

非正式地，odr-use 变量的含义如下:

If any expression anywhere in the program takes the address of or binds a reference directly to an object, this object must be defined.

最新草案中的说明

在最新版本的规范 §3.2 中已经阐明(见 Draft C++14 on GitHub):

2 An expression is potentially evaluated unless it is an unevaluated operand (Clause 5) or a subexpression thereof. The set of potential results of an expression e is defined as follows:

• If e is an id-expression (5.1.1), the set contains only e.
• If e is a class member access expression (5.2.5), the set contains the potential results of the object expression.
• If e is a pointer-to-member expression (5.5) whose second operand is a constant expression, the set contains the potential results of the object expression.
• If e has the form (e1), the set contains the potential results of e1.
• If e is a glvalue conditional expression (5.16), the set is the union of the sets of potential results of the second and third operands.
• If e is a comma expression (5.18), the set contains the potential results of the right operand.
• Otherwise, the set is empty.

[ Note: This set is a (possibly-empty) set of id-expressions, each of which is either e or a subexpression of e.

[ Example: In the following example, the set of potential results of the initializer of n contains the first S::x subexpression, but not the second S::x subexpression.

struct S { static const int x = 0; };
const int &f(const int &r);
int n = b ? (1, S::x) // S::x is not odr-used here
          : f(S::x);  // S::x is odr-used here, so
                      // a definition is required

—end example ] —end note ]

3 A variable x whose name appears as a potentially-evaluated expression ex is odr-used by ex unless applying the lvalue-to-rvalue conversion (4.1) to x yields a constant expression (5.19) that does not invoke any nontrivial 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 (4.1) is applied to e, or e is a discarded-value expression (Clause 5).

C++11 的情况如何？

C++11 中的§3.2/2 内容如下:

An expression is potentially evaluated unless it is an unevaluated operand (Clause 5) or a subexpression thereof. A variable whose name appears as a potentially-evaluated expression is odr-used unless it is an object that satisfies the requirements for appearing in a constant expression (5.19) and the lvalue-to-rvalue conversion (4.1) is immediately applied.

这些措辞的问题是DR 712 .考虑这个例子:

struct S {
  static const int a = 1;
  static const int b = 2;
};
int f(bool x) {
  return x ? S::a : S::b;
}

由于 S::a 和 S::b 是左值，条件表达式 x ? S::a : S::b 也是一个左值。这意味着左值到右值的转换不会立即应用于 S::a 和 S::b，而是应用于结果的条件表达式。这意味着按照 C++11 的措辞，这些静态数据成员是 odr-used 并且需要定义。但实际上只使用了值，因此不需要定义静态数据成员 - 声明就足够了。 C++14 草案的新措辞解决了这个问题。

新措辞是否解决了所有问题？

没有。在以下示例中，变量 S::a 仍然是 odr-used:

struct S { static constexpr int a[2] = {0, 1}; };
void f() {
    auto x = S::a[0];
}

因此我提交了一个新的 issue将以下项目符号添加到 §3.2/2:

• if e is a glvalue subscripting expression (5.2.1) of the form E1[E2], the set contains the potential results of E1.

关于c++ - 何时在 C++14 中使用变量 odr？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/23491781/