c++ - initializer_list::size() 上的 static_assert

Question

为什么是 std::initializer_list<_E>::size不允许在 static_assert 中使用，即使它被声明为 constexpr在我的 libstdc++ (v. 4.6) 中？

例如下面的代码:

template<class T, int Length>
class Point
{
  public:
    Point(std::initializer_list<T> init)
    {
      static_assert(init.size() == Length, "Wrong number of dimensions");
    }
};

int main()
{
  Point<int, 3> q({1,2,3});

  return 0;
}

给出以下错误:

test.C: In constructor ‘Point<T, Length>::Point(std::initializer_list<_Tp>) [with T = int, int Length = 3]’:
test.C:60:26:   instantiated from here
test.C:54:7: error: non-constant condition for static assertion
test.C:54:73:   in constexpr expansion of ‘init.std::initializer_list<_E>::size [with _E = int, std::initializer_list<_E>::size_type = long unsigned int]()’
test.C:54:7: error: ‘init’ is not a constant expression

请注意，这对于一个简单的示例来说效果很好:

class A
{
  public:
    constexpr int size() { return 5; }
};

int main()
{
  A a;
  static_assert(a.size() == 4, "oh no!");

  return 0;
}

Answer 1

“初始化列表”只是可怕的杂物。

不要:

#include <initializer_list>

template<typename T>
void Dont(std::initializer_list<T> list) { // Bad!
    static_assert(list.size() == 3, "Exactly three elements are required.");
}

void Test() { Dont({1,2,3}); }

做:

template<typename T, std::size_t N>
void Do(const T(&list)[N]) { // Good!
    static_assert(N == 3, "Exactly three elements are required.");
}

void Test() { Do({1,2,3}); }