constexpr 与 const 引用交互的奇怪行为

Question

我用 clang 和 gcc（主干版本）测试了以下代码。有人可以解释为什么使用普通 X 结构的情况不起作用，而按值捕获和使用 const 引用的包装器情况都可以正常工作。

struct X {
  constexpr X(const int &v) : x(v) {}
  constexpr int Get() const { return x; }
private:
  const int& x;
};

constexpr X f(const X& r) {
  return r;
}

struct Y {
  constexpr Y(const int &v) : x(v) {}
  constexpr int Get() const { return x; }
private:
  const int x;
};

constexpr Y f(const Y& r) {
  return r;
}

struct Wrap {
  constexpr Wrap(const int& v) : x(v) {}
  constexpr Y Get() const { return Y{x}; }
private:
  const int x;
};


int main() {
  constexpr const int x = 10;

  /* this does not work for some reason 
  constexpr X a(x);
  static_assert(f(a).Get() == 10, "This should work.");
  */

  // This works.
  constexpr Y b(x);
  static_assert(f(b).Get() == 10, "This should work.");

  // This also works.
  constexpr Wrap c(x);
  static_assert(f(c.Get()).Get() == 10, "This should work.");

  return 0;
}

Answer 1

您在这里违反的规则：

constexpr X a(x);

consetxpr 指针或 constexpr 引用是否必须引用具有静态存储持续时间的对象 - 这是其自身地址成为常量表达式的唯一方法。那是 而不是 x 的情况。

但是一旦你做到了（const 是多余的）：

static constexpr int x = 10;

然后剩下的工作：

constexpr X a(x);
static_assert(f(a).Get() == 10, "This should work.");

具体规则为[expr.const]/11:

A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints:

if the value is an object of class type, each non-static data member of reference type refers to an entity that is a permitted result of a constant expression,

[...]

An entity is a permitted result of a constant expression if it is an object with static storage duration that either is not a temporary object or is a temporary object whose value satisfies the above constraints, or if it is a non-immediate function.

另见 of 。我可能应该关闭所有这些作为其中之一的欺骗？

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