引用类型和 odr 使用的模板非类型参数

Question

下面示例代码中的变量v是否odr-used?

extern void* v;

template<void*&>
void f() {}

int main()
{
    f<v>();
}

我在 Boost ML 中发现了这个模式。

比照。 http://lists.boost.org/Archives/boost/2011/04/180082.php

它说 boost::enabler 从未定义，但如果提供 -g 选项，clang 会拒绝它作为链接错误。

比照。 http://melpon.org/wandbox/permlink/nF45k7un3rFb175z

上面的示例代码是 Boost ML 代码的简化版本，clang 也拒绝它。

比照。 http://melpon.org/wandbox/permlink/ZwxaygXgUhbi1Cbr

我认为（但我不确定）引用类型的模板非类型参数是 odr-used 即使它们没有在模板主体中引用所以 Boost ML 的模式格式错误。

我的理解正确吗？

Answer 1

我相信 v 是 odr-used。 f<v> 是一个 template-id (14.2) 其 template-argument 是一个 id-expression (5.1.1) - 一种表达形式。它显然不是未求值的操作数（它没有作为 typeid、sizeof、noexcept 或 decltype 的操作数出现），因此它是 可能求值的 根据 3.2/2:

3.2/2 An expression is potentially evaluated unless it is an unevaluated operand (Clause 5) or a subexpression thereof...

此时，我们有

3.2/3 A variable x whose name appears as a potentially-evaluated expression ex is odr-used unless [a condition that doesn't appear to apply here as no lvalue-to-rvalue conversion is applied].

Answer 2

[basic.def.odr]/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) [..]

不幸的是，此时将 l-t-r 转换应用于 v 不会产生常量表达式 - [expr.const]/2:

A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine (1.9), would evaluate one of the following expressions: [..]

— an lvalue-to-rvalue conversion (4.1) unless it is applied to

a non-volatile glvalue of integral or enumeration type that refers to a non-volatile const object with a preceding initialization, initialized with a constant expression [..], or

a non-volatile glvalue that refers to a non-volatile object defined with constexpr, or that refers to a non-mutable sub-object of such an object, or

a non-volatile glvalue of literal type that refers to a non-volatile object whose lifetime began within the evaluation of e;

然而，尽管 Matt 建议的实现方式不正确，但这个想法肯定是正确的。中演示了一种使用此方法的简单方法，即使用辅助模板。在你的情况下，试试

template <bool cond, int id=0>
using distinct_enable_if =
    typename std::enable_if<cond, std::integral_constant<int,id>*>::type;

class test
{
public:
  template< class... T,
            distinct_enable_if<sizeof...(T) == 10> = nullptr> 
  test( T&&... ) {}

  template< class T,
            distinct_enable_if<std::is_arithmetic<T>{}> = nullptr>
  operator T() const { return T{}; }

  /* Note the additional template argument:
     It ensures that the template parameter lists are not identical, 
     and the ODR isn't violated */
  template< class T,
            distinct_enable_if<std::is_pointer<T>{}, 1> = nullptr>
  operator T() const { return T{}; }
};

Demo.

引用类型和 odr 使用的模板非类型参数

Template non-type arguments for reference type and odr-used

c++

boost

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