相关疑难解决方法(0)

下面的代码合法吗？

template <int N>
class foo {
public:
    constexpr foo()
    {
        for (int i = 0; i < N; ++i) {
            v_[i] = i;
        }
    }

private:
    int v_[N];
};

constexpr foo<5> bar;

Clang 接受它，但 GCC 和 MSVC 拒绝它。

GCC 的错误是：

main.cpp:15:18: error: 'constexpr foo<N>::foo() [with int N = 5]' called in a constant expression
   15 | constexpr foo<5> bar;
      |                  ^~~
main.cpp:4:15: note: 'constexpr foo<N>::foo() [with int N = 5]' is not usable as a 'constexpr' function because:
    4 …

我想要一个constexpr可以为每个C ++类型返回唯一ID 的函数，如下所示：

using typeid_t = uintptr_t;

template <typename T>
constexpr typeid_t type_id() noexcept
{
  return typeid_t(type_id<T>);
}

int main()
{
  ::std::cout << ::std::integral_constant<typeid_t, type_id<float>()>{} << ::std::endl;

  return 0;
}

但是我得到一个错误：

t.cpp: In function 'int main()':
t.cpp:23:69: error: conversion from pointer type 'typeid_t (*)() noexcept {aka long unsigned int (*)() noexcept}' to arithmetic type 'typeid_t {aka long unsigned int}' in a constant-expression
   ::std::cout << ::std::integral_constant<typeid_t, type_id<float>()>{} << ::std::endl;
                                                                     ^
t.cpp:23:69: note: in template argument for type 'long unsigned …