c++ - GCC 4.9 constexpr 错误的解决方法

Question

我有以下代码片段，它代表了实际更大的代码片段:

#include <iostream>
using namespace std;

template<size_t N> class A {
    public:
        static constexpr size_t getN() {return N;}
};

template<size_t N> class B {
    public:
        void print() { cout << "B created: " << N << '\n';}
};

template <class T> class C {
    public:
        void set(T* a) {
            t_ptr = a;
        }

        void create() {
            constexpr int m = t_ptr->getN();
            B<m> b;
            b.print();
        }

    private:
        T* t_ptr;
};

int main() {
   constexpr int n = 2;
   A<n> a;
   C<A<n> > c;
   c.set(&a);
   c.create();
}

使用 g++ -o main main.cpp -std=c++11 和 GCC/G++ 4.8.3 进行编译，收到预期输出: B 创建:2

但是，使用 GCC/G++ 4.9.1 代码无法编译，输出:

main.cpp: In member function ‘void C<T>::create()’:
main.cpp:27:15: error: the value of ‘m’ is not usable in a constant expression
             B<m> b;
               ^
main.cpp:26:27: note: ‘m’ used in its own initializer
             constexpr int m = t_ptr->getN();
                           ^
main.cpp:27:16: error: the value of ‘m’ is not usable in a constant expression
             B<m> b;
                ^
main.cpp:26:27: note: ‘m’ used in its own initializer
             constexpr int m = t_ptr->getN();
                           ^
main.cpp:27:19: error: invalid type in declaration before ‘;’ token
             B<m> b;
                   ^
main.cpp:28:15: error: request for member ‘print’ in ‘b’, which is of non-class type ‘int’
             b.print();
               ^

这是由 GCC 4.9 中的一个已知错误引起的:https://gcc.gnu.org/bugzilla/show_bug.cgi?id=59937在这个旧线程中 https://gcc.gnu.org/ml/gcc-bugs/2013-11/msg00067.html建议使用 extern 作为解决方法。但是，我无法使此解决方法发挥作用。

你们能帮我让这段代码在 GCC 4.9 中编译吗？谢谢!

Answer 1

由于 this 不是 constexpr，因此对 this->t_ptr 的访问也不是。

clang 的错误更有帮助

implicit use of 'this' pointer is only allowed within the
    evaluation of a call to a 'constexpr' member function

引用:

N3690 5.19/2 (emphasis added)

A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

— this, except in a constexpr function or a constexpr constructor that is being evaluated as part of e;

通过类型名调用静态成员函数是可行的

constexpr int m = T::getN();